Probability ML Aggarwal Solutions ICSE Class-10 Maths

Chapter-22 Probability APC Understanding Mathematics Solutions Exe-22

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Probability ML Aggarwal Solutions ICSE Class-10 Maths Chapter-22 . We Provide Step by Step Answer of Exercise-22,  with MCQs and Chapter-Test of Probability Questions  / Problems related  for ICSE Class-10 APC Understanding Mathematics  . Visit official Website CISCE  for detail information about ICSE Board Class-10.

Probability ML Aggarwal Solutions ICSE Class-10 Maths Chapter-22


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Exercise-22 ML Aggarwal Solutions Probability ICSE  Class 10

Question 1

A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. Anjali takes out a ball from the bag without looking into it. What is the probability that she takes out
(i) yellow ball ?
(ii) red ball ?
(iii) blue ball ?

Answer 1

Number of balls in the bag = 3.
(i) Probability of yellow ball = \\ \frac { 1 }{ 3 }
(ii) Probability of red ball = \\ \frac { 1 }{ 3 }
(iii) Probability of blue ball = \\ \frac { 1 }{ 3 }

Question 2

A box contains 600 screws, one-tenth are rusted. One screw is taken out at random from this box. Find the probability that it is a good screw.

Answer 2

Number of total screws = 600
Rusted screws = \\ \frac { 1 }{ 10 }  of 600 = 60
∴ Good screws = 600 – 60 = 540
Probability of a good screw
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 540 }{ 600 }
\\ \frac { 9 }{ 10 }

Question 3

In a lottery, there are 5 prized tickets and 995 blank tickets. A person buys a lottery ticket. Find the probability of his winning a prize.

Answer 3

Number of prized tickets = 5
Number of blank tickets = 995
Total number of tickets = 5 + 995 = 1000
Probability of prized ticket
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 5 }{ 1000 }
\\ \frac { 1 }{ 200 }

Question 4

12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Answer 4

Number of defective pens = 12
Number of good pens = 132
Total number of pens =12 + 132 = 144
Probability of good pen
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 132 }{ 144 }
\\ \frac { 11 }{ 12 }

Question 5

If the probability of winning a game is \\ \frac { 5 }{ 11 } , what is the probability of losing ?

Answer 5

Probability of winning game = \\ \frac { 5 }{ 11 }
⇒ P(E) = \\ \frac { 5 }{ 11 }
We know that P (E) + P (\overline { E } ) = 1
where P (E) is the probability of losing the game.
\\ \frac { 5 }{ 11 }  + P (\overline { E } ) = 1
⇒ P (\overline { E } ) = 1- \frac { 5 }{ 11 }
\\ \frac { 11-5 }{ 11 }
\\ \frac { 6 }{ 11 }

Question 6

Two players, Sania and Sonali play a tennis match. It is known that the probability of Sania winning the match is 0.69. What is the probability of Sonali winning ?

Answer 6

Probability of Sania’s winning the game = 0.69
Let P (E) be the probability of Sania’s winning the game
and P (\overline { E } ) be the probability of Sania’s losing
the game or probability of Sonali, winning the game
P (E) + P (\overline { E } ) = 1
⇒ 0.69 + P (\overline { E } ) = 1
⇒ P(\overline { E } ) = 1 – 0.69 = 0.31
Hence probability of Sonali’s winning the game = 0.31

Question 7

A bag contains 3 red balls and 5 black balls. A ball is drawn at random’ from in bag. What is the probability that the ball drawn is .
(i) red ?
(ii) not red ?

Answer 7

Number of red balls = 3
Number of black balls = 5
Total balls = 3 + 5 = 8
Let P (E) be the probability of red balls,
then P (\overline { E } ) will be the probability of not red balls.
P (E) + P (\overline { E } ) = 1
(i) But P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 3 }{ 8 }
(ii) P (\overline { E } ) = 1 – P(E)
1- \frac { 6 }{ 11 }
\\ \frac { 8-3 }{ 8 }
\\ \frac { 5 }{ 8 }

Question 8

There are 40 students in Class X of a school of which 25 are girls and the.others are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of
(i) a girl ?
(ii) a boy ?

Answer 8

Number of total students = 40
and Number of girls = 25
while Number of boys = 40 – 25 = 15
(i) Probability of a girl
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 25 }{ 40 }
\\ \frac { 5 }{ 8 }
(ii) Probability of a boy
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 15 }{ 40 }
\\ \frac { 3 }{ 8 }

Question 9

A letter is chosen from the word ‘TRIANGLE’. What is the probability that it is a vowel ?

Answer 9

There are three vowels: I, A, E
.’. The number of letters in the word ‘TRIANGLE’ = 8.
Probability of vowel
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 3 }{ 8 }

Question 10

A letter of English alphabet is chosen at random. Determine the probability that the letter is a consonant.

Answer 10

No. of English alphabet = 26
and No. of vowel = 5
while No. of constant = 25 – 5 = 21
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 21 }{ 26 }

Question 11

A bag contains 5 black, 7 red and 3 white balls. A ball is drawn at random from the bag, find the probability that the ball drawn is:
(i) red
(ii) black or white
(iii) not black.

Answer 11

In a bag,
Number of black balls = 5
and Number of red balls = 7
while  number of white balls = 3
Total number of balls in the bag
= 5 + 7 + 3 = 15
(i) Probability of red balls
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 7 }{ 15 }

(ii) Probability of black or white balls

P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 5+3 }{ 15 }
\\ \frac { 8 }{ 15 }
(iii) Probability of not black balls
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 7+3 }{ 15 }
\\ \frac { 10 }{ 15 }
\\ \frac { 2 }{ 3 }

Question 12

A box contains 7 blue, 8 white and 5 black marbles. If a marble is drawn at random from the box, what is the probability that it will be
(i) black?
(ii) blue or black?
(iii) not black?
(iv) green?

Answer 12

Total number of marbles in the box
= 7 + 8 + 5 = 20
Since, a marble is drawn at random from the box

(i) Probability (of a black Marble)

P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 5 }{ 20 }
\\ \frac { 1 }{ 4 }

(ii) Probability (of a blue or black marble)

P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 7+5 }{ 20 }
and = \\ \frac { 12 }{ 20 }
hence = \\ \frac { 3 }{ 5 }

(iii) Probability (of not black marble)

= 1 – P (of black 1)
and = 1- \frac { 1 }{ 4 }
so = \\ \frac { 4-1 }{ 4 }
Hence = \\ \frac { 3 }{ 4 }
(iv) P (of a green marble) = 0
(∴ Since, a box does not contain a green marble,
so the probability of green marble will be zero)

Question 13

A bag contains 6 red balls, 8 white balls, 5 green balls and 3 black balls. One ball is drawn at random from the bag. Find the probability that the ball is :
(i) white
(ii) red or black
(iii) not green
(iv) neither white nor black.

Answer 13

In a bag,
Number of red balls = 6
and Number of white balls = 8
while Number of green balls = 5
and number of black balls = 3
Total number of balls in the bag
= 6 + 8 + 5 + 3 = 22

(i) Probability of white balls

P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 8 }{ 22 }
\\ \frac { 4 }{ 11 }

(ii) Probability of red or black balls

P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 6+3 }{ 22 }
\\ \frac { 9 }{ 22 }

(iii) Probability of not green balls i.e. having red, white and black balls.

P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 6+8+3 }{ 22 }
\\ \frac { 17 }{ 22 }

(iv) Probability of neither white nor black balls i.e. red and green balls

P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 6+5 }{ 22 }
\\ \frac { 11 }{ 22 }
\\ \frac { 1 }{ 2 }

Question 14

A piggy bank contains hundred 50 p coins, fifty Rs 1 coins, twenty Rs 2 coins and ten Rs 5 coins. It is equally likely that one of the coins will fall down when the bank is turned upside down, what is the probability that the coin
(i) will be a 50 p coin?
(ii) will not be Rs 5 coin?

Answer 14

In a piggy bank, there are
100, 50 p coin
50, Rs 1 coin
20, Rs 2 coin
10, Rs 5 coin
Total coins = 100 + 50 + 20 + 10 = 180
One coin is drawn at random Probability of
(i) 50 p coins = \\ \frac { 100 }{ 180 }
\\ \frac { 5 }{ 9 }
(ii) Will not be Rs 5 coins
= 100 + 50 + 20 = 170
Probability = \\ \frac { 170 }{ 180 }  = \\ \frac { 17 }{ 18 }

Question 15

A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Peter, a trader, will only accept the shirts which are good, but Salim, another trader, will only reject the shirts which have major defects. One shirts is drawn at random from the carton. What is the probability that
(i) it is acceptable to Peter ?
(ii) it is acceptable to Salim ?

Answer 15

In a carton, there the 100 shirts.
Among these number of shirts which are good = 88
number of shirts which have minor defect = 8
number of shirt which have major defect = 4
Total number of shirts = 88 + 8 + 4 = 100
Peter accepts only good shirts i.e. 88
Salim rejects only shirts which have major defect i.e. 4

(i) Probability of good shirts which are acceptable to Peter

P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 88 }{ 100 }
\\ \frac { 22 }{ 25 }
(ii) Probability of shirts acceptable to Salim
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 88+8 }{ 100 }
\\ \frac { 96 }{ 100 }
\\ \frac { 24 }{ 25 }

Question 16

A die is thrown once. What is the probability that the
(i) number is even
(ii) number is greater than 2 ?

Answer 16

Dice is thrown once
Sample space = {1, 2, 3, 4, 5, 6}
(i) No. of ways in favour = 3
(∵ Even numbers are 2, 4, 6)
Total ways = 6
Probability = \\ \frac { 3 }{ 6 }  = \\ \frac { 1 }{ 2 }
(ii) No. of ways in favour = 4
(Numbers greater than 2 are 3, 4, 5, 6)
Total ways = 6
Probability = \\ \frac { 4 }{ 6 }  = \\ \frac { 2 }{ 2 }

Question 17

In a single throw of a die, find the probability of getting:
(i) an odd number
(ii) a number less than 5
(iii) a number greater than 5
(iv) a prime number
(v) a number less than 8
(vi) a number divisible by 3
(vii) a number between 3 and 6
(viii) a number divisible by 2 or 3.

Answer 17

A die is thrown and on its faces, numbers 1 to 6 are written.
Total numbers of possible outcomes = 6
(i) Probability of an odd number,
odd number are 1, 3 and 5
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 3 }{ 6 }
\\ \frac { 1 }{ 2 }

(ii) A number less them 5 are 1, 2, 3, 4

Probability of a number less than 5 is
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 4 }{ 6 }
\\ \frac { 2 }{ 3 }

(iii) A number greater than 5 is 6

Probability of a number greater than 5 is
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 1 }{ 6 }
(iv) Prime number is 2, 3, 5
Probability of a prime number is
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 3 }{ 6 }
\\ \frac { 1 }{ 2 }

(v) Number less than 8 is nil

P (E) = 0
(vi) A number divisible by 3 is 3, 6
Probability of a number divisible by 3 is
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 2 }{ 6 }
\\ \frac { 1 }{ 3 }
(vii) Numbers between 3 and 6 is 4, 5
Probability of a number between 3 and 6 is
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 2 }{ 6 }
\\ \frac { 1 }{ 3 }

(viii) Numbers divisible by 2 or 3 are 2, 4 or 3,

Probability of a number between 2 or 3 is
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 2 }{ 6 }
\\ \frac { 1 }{ 3 }

Question 18

A die has 6 faces marked by the given numbers as shown below:
\boxed { \quad 1 } \boxed { \quad 2 } \boxed { \quad 3 } \boxed { -1 } \boxed { -2 } \boxed { -3 }
The die is thrown once. What is the probability of getting
(i) a positive integer.
(ii) an integer greater than – 3.
(iii) the smallest integer ?

Answer 18

Total outcomes n(S)= 6
(i) a positive integer = (1, 2, 3)
No. of favourables n(E) = 3
Probability = \\ \frac { n(E) }{ n(S) }
\\ \frac { 3 }{ 6 }
\\ \frac { 1 }{ 2 }

(ii) Integer greater than -3

= (1, 2, 3, -1, -2)
No. of favourables n(E) = 5
Probability = \\ \frac { n(E) }{ n(S) }
\\ \frac { 5 }{ 6 }
(iii) Smallest integer = -3
No. of favourables n(E) = 1
Probability = \\ \frac { n(E) }{ n(S) }
\\ \frac { 1 }{ 6 }

Question 19

A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (shown in the adjoining figure) and these are equally likely outcomes. What is the probability that it will point at

(i) 8 ?
(ii) an odd number ?
(iii) a number greater than 2?
(iv) a number less than 9?
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Ex 22 19
Answer 19

On the face of a game, numbers 1 to 8 is shown.
Possible outcomes = 8
(i) Probability of number 8 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 1 }{ 8 }
(ii) Odd number are 1, 3, 5, 7
Probability of a number which is an odd will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 4 }{ 8 }
\\ \frac { 1 }{ 2 }

(iii) A number greater than 2 are 3, 4, 5, 6, 7, 8 which are 6

Probability of number greater than 2 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 6 }{ 8 }
\\ \frac { 3 }{ 4 }
(iv) A number less than 9 is 8.
Probability of a number less than 9 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 8 }{ 8 }

Question 20

Find the probability that the month of January may have 5 Mondays in
(i) a leap year
(ii) a non-leap year.

Answer 20

In January, there are 31 days and in an ordinary year,
there are 365 days but in a leap year, there are 366 days.

(i) In January of an ordinary year, there are 31 days i.e. 4 weeks and 3 days.
Probability of Monday will be = \\ \frac { 3 }{ 7 }

(ii) In January of a leap year, there are 31 days i.e. 4 weeks and 3 days
Probability of Monday will be = \\ \frac { 3 }{ 7 }

Question 21

Find the probability that the month of February may have 5 Wednesdays in
(i) a leap year
(ii) a non-leap year.

Answer 21

In the month of February, there are 29 days in a leap year
while 28 days in a non-leap year,
(i) In a leap year, there are 4 complete weeks and 1 day
Probability of Wednesday = P (E) = \\ \frac { 1 }{ 7 }
(ii) and in a non leap year, there are 4 complete weeks and 0 days
Probability of Wednesday P (E) = \\ \frac { 0 }{ 7 }  = 0

Question 22

Sixteen cards are labelled as a, b, c,…, m, n, o, p. They are put in a box and shuffled. A boy is asked to draw a card from the box. What is the probability that the card drawn is:
(i) a vowel
(ii) a consonant
(iii) none of the letters of the word median.

Answer 22

Here, sample space (S) = {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p)
∴n(S) = 16
(i) Vowels (V) = {a, e, i, o}
∴n(V) = 4
∴P(a vowel) = \\ \frac { n(V) }{ n(S) }  = \\ \frac { 4 }{ 16 }  = \\ \frac { 1 }{ 4 }

(ii) Consonants (C) = {b, c, d, f, g, h, j, k, l, m, n, p}

∴n(C) = 12
∴P (a consonant) = \\ \frac { n(C) }{ n(S) }  = \\ \frac { 12 }{ 16 }  = \\ \frac { 3 }{ 4 }
(iii) None of the letters of the word MEDIAN (N) = {b, c, f, g, h, j, k, l, o, p)
∴n(N) = 10
∴P (N) = \\ \frac { n(N) }{ n(S) }  = \\ \frac { 10 }{ 16 }  = \\ \frac { 5 }{ 8 }

Question 23

An integer is chosen between 0 and 100. What is the probability that it is
(i) divisible by 7?
(ii) not divisible by 7?

Answer 23

Integers between 0 and 100 = 99
(i) Number divisible by 7 are
7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98 = 14
Probability = \\ \frac { 14 }{ 99 }
(ii) Not divisible by 7 are 99 – 14 = 85
Probability = \\ \frac { 85 }{ 99 }

Question 24

Cards marked with numbers 1, 2, 3, 4, 20 are well shuffled and a card is drawn at random.
What is the probability that the number on the card is
(i) a prime number
(ii) divisible by 3
(iii) a perfect square ? (2010)

Answer 24

Number cards is drawn from 1 to 20 = 20
One card is drawn at random
No. of total (possible) events = 20
(i) The card has a prime number
The prime number from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19
Actual No. of events = 8
P(E) = \frac { Number\quad of\quad actual\quad events }{ Number\quad of\quad total\quad events }
\\ \frac { 8 }{ 20 }
\\ \frac { 2 }{ 5 }

(ii) Numbers divisible by 3 are 3, 6, 9, 12, 15, 18

No. of actual events = 6
P(E) = \frac { Number\quad of\quad actual\quad events }{ Number\quad of\quad total\quad events }
\\ \frac { 6 }{ 20 }
\\ \frac { 3 }{ 10 }
(iii) Numbers which are perfect squares = 1, 4, 9, 16 = 4
P(E) = \frac { Number\quad of\quad actual\quad events }{ Number\quad of\quad total\quad events }
\\ \frac { 4 }{ 20 }
\\ \frac { 1 }{ 5 }

Question 25

A box contains 25 cards numbered 1 to 25. A card is drawn from the box at random. Find the probability that the number on the card is :
(i) even
(ii) prime
(iii) multiple of 6

Answer 25

Number of card in a box = 25 numbered 1 to 25
(i) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24
i.e. number of favourable outcomes = 12
Probability of an even number will be
P(E) = \\ \frac { 12 }{ 25 }

(ii) Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23

i.e. number of primes = 9
Probability of primes will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 9 }{ 25 }
(iii) Multiples of 6 are 6, 12, 18, 24
Number of multiples = 4
Probability of multiples of 6 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 4 }{ 25 }

Question 26

A box contains 15 cards numbered 1, 2, 3,…..15 which are mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the card is :
(i) Odd
(ii) prime
(iii) divisible by 3
(iv) divisible by 3 and 2 both
(v) divisible by 3 or 2
(vi) a perfect square number.

Answer 26

Number of cards in a box =15 numbered 1 to 15

(i) Odd numbers are 1, 3, 5, 7, 9, 11, 13, 15

Number of odd numbers = 8
Probability of odd numbers will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 8 }{ 15 }

(ii) Prime number are 2, 3, 5, 7, 11, 13

Number of primes is 6
Probability of prime number will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 6 }{ 15 }
\\ \frac { 2 }{ 5 }

(iii) Numbers divisible by 3 are 3, 6, 9, 12, 15

which are 5 in numbers
Probability of number divisible by 3 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 5 }{ 15 }
\\ \frac { 1 }{ 3 }

(iv) Divisible by 3 and 2 both are 6, 12

which are 2 in numbers.
Probability of number divisible by 3 and 2
Both will be = \\ \frac { 2 }{ 15 }

(v) Numbers divisible by 3 or 2 are

2, 3, 4, 6, 8, 9, 10, 12, 14, 15 which are 10 in numbers
Probability of number divisible by 3 or 2 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 10 }{ 15 }
\\ \frac { 2 }{ 3 }

(v) Perfect squares number are 1, 4, 9 i.e., 3 number

P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 3 }{ 15 }
\\ \frac { 1 }{ 5 }

Question 27

A box contains 19 balls bearing numbers 1, 2, 3,…., 19. A ball is drawn at random
from the box. Find the probability that the number on the ball is :
(i) a prime number
(ii) divisible by 3 or 5
(iii) neither divisible by 5 nor by 10
(iv) an even number.

Answer 27

In a box, number of balls = 19 with number 1 to 19.
A ball is drawn
Number of possible outcomes = 19

(i) Prime number = 2, 3, 5, 7, 11, 13, 17, 19

which are 8 in number
Probability of prime number will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 8 }{ 19 }

(ii) Divisible by 3 or 5 are 3, 5, 6, 9, 10, 12, 15, 18

which are 8 in number
Probability of number divisible by 3 or 5 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 8 }{ 19 }

(iii) Numbers which are neither divisible by 5 nor by 10 are

1, 2, 3, 4, 6, 7, 8, 9, 11, 12,
13, 14, 16, 17, 18, 19
which are 16 in numbers
Probability of there number will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 16 }{ 19 }

(iv) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18

which are 9 in numbers.
Probability of there number will be
Number of favourable outcome
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 9 }{ 19 }

Question 28

Cards marked with numbers 13, 14, 15, …, 60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card drawn is
(i) divisible by 5

(ii) a perfect square number.

Answer 28

Number of card bearing numbers 13,14,15, … 60 = 48
One card is drawn at random.
(i) Card divisible by 5 are 15, 20, 25, 30, 35, 40, 45, 50, 55, 60 = 10
Probability = \\ \frac { 10 }{ 48 }
\\ \frac { 5 }{ 24 }
(ii) A perfect square = 16, 25, 36, 49 = 4
Probability = \\ \frac { 4 }{ 48 }
\\ \frac { 1 }{ 12 }

Question 29

Tickets numbered 3, 5, 7, 9,…., 29 are placed in a box and mixed thoroughly. One ticket is drawn at random from the box. Find the probability that the number on the ticket is
(i) a prime number
(ii) a number less than 16
(iii) a number divisible by 3.

Answer 29

In a box there are 14 tickets with number
3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29
Number of possible outcomes = 14
(i) Prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29
which are 9 in number
Probability of prime will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 9 }{ 14 }
(ii) Number less than 16 are 3, 5, 7, 9, 11, 13, 15
which are 7 in numbers,
Probability of number less than 16 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 7 }{ 14 }
\\ \frac { 1 }{ 2 }
(iii) Numbers divisible by 3 are 3, 9, 15, 21, 27
which are 5 in number
Probability of number divisible by 3 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 5 }{ 14 }

Question 30

A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5.

Answer 30

There are 90 discs in a box containing numbered from 1 to 90.
Number of possible outcomes = 90
(i) Two digit numbers are 10 to 90 which are 81 in numbers.
Probability of two digit number will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 81 }{ 90 }
\\ \frac { 9 }{ 10 }
(ii) Perfect squares are 1, 4, 9, 16, 25, 36,49, 64, 81
which are 9 in numbers.
Probability of square will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 9 }{ 90 }
\\ \frac { 1 }{ 10 }
(iii) Number divisible by 5 are
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90
which are 18 in numbers.
Probability of number divisible by 5 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 18 }{ 90 }
\\ \frac { 1 }{ 5 }

Question 31

Cards marked with numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn at random from this box. Find the probability that the number on the card is
(i) an even number
(ii) a number less than 14
(iii) a number which is a perfect square
(iv) a prime number less than 30.

Answer 31

Number of cards with numbered from 2 to 101 are placed in a box
Number of possible outcomes = 100 one card is drawn
(i) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16,….., 96, 98, 100
which are 50 in numbers.
Probability of even number will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 50 }{ 100 }
\\ \frac { 1 }{ 2 }
(ii) Numbers less than 14 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13
which are 12 in numbers
Probability of number less than 14 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 12 }{ 100 }
\\ \frac { 3 }{ 25 }
(iii) Perfect square are 4, 9, 16, 25, 36, 49, 64, 81, 100 which are 9 in numbers
Probability of perfect square number will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 9 }{ 100 }
(iv) Prime numbers less than 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
which are 10 in numbers Probability of prime numbers, less than 30 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 10 }{ 100 }
\\ \frac { 1 }{ 10 }

Question 32

A bag contains 15 balls of which some are white and others are red. If the probability of drawing a red ball is twice that of a white ball, find the number of white balls in the bag.

Answer 32

In a bag, there are 15 balls.
Some are white and others are red.
Probability of red ball = 2 probability of white ball
Let number of white balls = x
Then, number of red balls = 15 – x
2\times \frac { 15-x }{ 15 } =\frac { x }{ 15 }
⇒ 2(15 – x) = x
and ⇒ 30 – 2x = x
so ⇒ 30 = x + 2x
Hence ⇒ x = \\ \frac { 30 }{ 3 }  = 10
Number of red balls = 10
and Number of white balls = 15 – 10 = 5

Question 33

A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball is twice that of a red ball, find the number of balls in the bag.

Answer 33

In a bag, there are 6 red balls, and some blue balls
Probability of blue ball = 2 × probability of red ball
Let number of blue balls = x
and number of red balls = 6
Total balls = x + 6
Probability of a blue ball = 2
⇒ \frac { x }{ x+6 } =2\times \frac { 6 }{ x+6 }
and ⇒ \frac { x }{ x+6 } =\frac { 12 }{ x+6 }
Hence ⇒ x = 12
Number of balls = x + 6 = 12 + 6 = 18

Question 34

A bag contains 24 balls of which x are red, 2x are white and 3x are blue. A blue is selected at random. Find the probability that it is
(i) white
(ii) not red.

Answer 34

In a bag, there are 24 balls
Since, there are x balls red, 2 × balls white and 3 × balls blue
x + 2x + 3x = 24
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Ex 22 34

Question 35

A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting:
(i) ‘2’ of spades
(ii) a jack .
(iii) a king of red colour
(iv) a card of diamond
(v) a king or a queen
(vi) a non-face card
(vii) a black face card
(viii) a black card
(ix) a non-ace
(x) non-face card of black colour
(xi) neither a spade nor a jack
(xii) neither a heart nor a red king

Answer 35

In a playing card, there are 52 cards
Number of possible outcome = 52

(i) Probability of‘2’ of spade will be

P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }

\\ \frac { 1 }{ 52 }

(ii) There are 4 jack card Probability of jack will be

P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 4 }{ 52 }
\\ \frac { 1 }{ 13 }

(iii) King of red colour are 2 in number

Probability of red colour king will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 2 }{ 52 }
\\ \frac { 1 }{ 26 }

(iv) Cards of diamonds are 13 in number

Probability of diamonds card will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 13 }{ 52 }
\\ \frac { 1 }{ 4 }

(v) Number of kings and queens = 4 + 4 = 8

P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 8 }{ 52 }
\\ \frac { 2 }{ 13 }

(vi) Non-face cards are = 52 – 3 × 4 = 52 – 12 = 40

Probability of non-face card will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 40 }{ 52 }
\\ \frac { 10 }{ 13 }

(vii) Black face cards are = 2 × 3 = 6

Probability of black face card will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 6 }{ 52 }
\\ \frac { 3 }{ 26 }

(viii) No. of black cards = 13 x 2 = 26

Probability of black card will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 26 }{ 52 }
\\ \frac { 1 }{ 2 }

(ix) Non-ace cards are 12 × 4 = 48

Probability of non-ace card will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 48 }{ 52 }
\\ \frac { 12 }{ 13 }

(x) Non-face card of black colours are 10 × 2 = 20

Probability of non-face card of black colour will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 20 }{ 52 }
\\ \frac { 5 }{ 13 }

(xi) Number of card which are neither a spade nor a jack

= 13 × 3 – 3 = 39 – 3 = 36
Probability of card which is neither a spade nor a jack will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 36 }{ 52 }
\\ \frac { 9 }{ 13 }

(xii) Number of cards which are neither a heart nor a red king

= 3 × 13 = 39 – 1 = 38
Probability of card which is neither a heart nor a red king will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 38 }{ 52 }
\\ \frac { 19 }{ 26 }

Question 36

All the three face cards of spades are removed from a well-shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting
(i) a black face card
(ii) a queen
(iii) a black card
(iv) a heart
(v) a spade
(vi) ‘9’ of black colour

Answer 36

In a pack of 52 cards
All the three face cards of spade are = 3
Number of remaining cards = 52 – 3 = 49
One card is drawn at random
(i) Probability of a black face card which are = 6 – 3 = 3
Probability = \\ \frac { 3 }{ 49 }
(ii) Probability of being a queen which are 4 – 1 = 3
Probability = \\ \frac { 3 }{ 49 }

(iii) Probability of being a black card = (26 – 3 = 23)

Probability = \\ \frac { 23 }{ 49 }
(iv) Probability of being a heart = \\ \frac { 13 }{ 49 }
(v) Probability of being a spade = (13 – 3 = 10)
Probability = \\ \frac { 10 }{ 49 }
(vi) Probability of being 9 of black colour (which are 2) = \\ \frac { 2 }{ 49 }

Question 37

From a pack of 52 cards, a blackjack, a red queen and two black kings fell down. A card was then drawn from the remaining pack at random. Find the probability that the card drawn is
(i) a black card
(ii) a king
(iii) a red queen.

Answer 37

In a pack of 52 cards, a blackjack, a red queen, two black being felt down.
Then number of total out comes = 52 – (1 + 1 + 2) = 48
(i) Probability of a black card (which are 26 – 3 = 23) = \\ \frac { 23 }{ 48 }
(ii) Probability of a being (4 – 2 = 2) = \\ \frac { 2 }{ 48 }  = \\ \frac { 1 }{ 24 }
(iii) Probability of a red queen = (2 – 1 = 1) = \\ \frac { 1 }{ 48 }

Question 38

Two coins are tossed once. Find the probability of getting:
(i) 2 heads
(ii) at least one tail.

Answer 38

Total possible outcomes are . HH, HT, TT, TH, i.e., 4
(i) Favourable outcomes are HH, i.e., 1
So, P(2 heads)
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 1 }{ 4 }
(ii) Favourable outcomes are HT, TT, TH, i.e., 3
So, P (at least one tail) = \\ \frac { 3 }{ 4 }

Question 39

Two different coins are tossed simultaneously. Find the probability of getting :
(i) two tails
(ii) one tail
(iii) no tail
(iv) atmost one tail.

Answer 39

Two different coins are tossed simultaneously
Number of possible outcomes = (2)² = 4
Number of event having two tails = 1 i.e. (T, T)
(i) Probability of two tails will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 1 }{ 4 }

(ii) Number of events having one tail = 2 i.e. (TH) and (HT)

Probability of one tail will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }

\\ \frac { 1 }{ 4 }

(iii) Number of events having no tail = 1 i.e. (HH)

Probability of having no tail will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 1 }{ 4 }
(iv) Atmost one tail
Number Of events having at the most one tail = 3 i.e. (TH), (HT, (TT)
Probability of at the most one tail will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 3 }{ 4 }

Question 40.

Two different dice are thrown simultaneously. Find the probability of getting:
(i) a number greater than 3 on each dice
(ii) an odd number on both dice.

Answer 40

When two different dice are thrown simultaneously,
then the sample space S of the random experiment =
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) .
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
It consists of 36 equally likely outcomes.

(i) Let E be the event of ‘a number greater than 3 on each dice’.

E = {(4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
No. of favorable outcomes (E) = 9
P (number greater than 3 on each dice) = \\ \frac { 9 }{ 36 }  = \\ \frac { 1 }{ 4 }

(ii) Let E be the event of ‘an odd number on both dice’.

E = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}
No. of favourable outcomes (E) = 9
∴ P (Odd on both dices) = \\ \frac { 9 }{ 36 }  = \\ \frac { 1 }{ 4 }

Question 41

Two different dice are thrown at the same time. Find the probability of getting :
(i) a doublet
(ii) a sum of 8
(iii) sum divisible by 5
(iv) sum of at least 11.

Answer 41

Two different dice are thrown at the same time
Possible outcomes will be (6)² i.e. 36

(i) Number of events which doublet = 6

i.e. (1, 1), (2, 2) (3, 3), (4, 4), (5, 5) and (6, 6)
.’. Probability of doublets will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 6 }{ 36 }
\\ \frac { 1 }{ 6 }

(ii) Number of event in which the sum is 8 are

(2, 6), (3, 5), (4, 4), (5, 3), (6, 2) = 5
Probability of a sum of 8 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 5 }{ 36 }

(iii) Number of event when sum is divisible by

5 are (1, 4), (4, 1), (2, 3), (3, 2), (4, 6),
(5, 5) = 7 in numbers
Probability of sum divisible by 5 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 7 }{ 36 }

(iv) Sum of atleast 11, will be in following events

(5, 6), (6, 5), (6, 6)
Probability of sum of atleast 11 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 3 }{ 36 }
\\ \frac { 1 }{ 12 }


MCQ Chapter-22 Probability ML Aggarwal Solutions for ICSE Board Class 10

Multiple Choice Questions

Choose the correct answer from the given four options (1 to 28):

Question 1

Which of the following cannot be the probability of an event?
(a) 0.7
(b) \\ \frac { 2 }{ 3 }
(c) – 1.5
(d) 15%

Answer 1

– 1.5 (negative) can not be a probability as a probability is possible 0 to 1. (c)

Question 2

If the probability of an event is p, then the probability of its complementary event will be
(a) p – 1
(b) p
(c) 1 – p
(d) 1- \frac { 1 }{ p }

Answer 2

Complementary of p is 1 – p
Probability of complementary even of p is 1 – p. (c)

Question 3

Out of one digit prime numbers, one selecting an even number is
(a) \\ \frac { 1 }{ 2 }
(b) \\ \frac { 1 }{ 4 }
(c) \\ \frac { 4 }{ 9 }
(d) \\ \frac { 2 }{ 5 }

Answer 3

One digit prime numbers are 2, 3, 5, 7 = 4
Probability of an even prime number (i.e , 2) = \\ \frac { 1 }{ 4 }  (b)

Question 4

Out of vowels, of the English alphabet, one letter is selected at random. The probability of selecting ‘e’ is
(a) \\ \frac { 1 }{ 26 }
(b) \\ \frac { 5 }{ 26 }
(c) \\ \frac { 1 }{ 4 }
(d) \\ \frac { 1 }{ 5 }

Answer 4

Vowels of English alphabet are a, e, i, o, u = 4
One letter is selected at random.
The probability of selecting ’e’ = \\ \frac { 1 }{ 5 }  (d)

Question 5

When a die is thrown, the probability of getting an odd number less than 3 is
(a) \\ \frac { 1 }{ 6 }
(b) \\ \frac { 1 }{ 3 }
(c) \\ \frac { 1 }{ 2 }
(d) 0

Answer 5

A die is thrown
Total number of events = 6
Odd number less than 3 is 1 = 1
Probability = \\ \frac { 1 }{ 6 }  (a)

Question 6

A fair die is thrown once. The probability of getting an even prime number is
(a) \\ \frac { 1 }{ 6 }
(b) \\ \frac { 2 }{ 3 }
(c) \\ \frac { 1 }{ 3 }
(d) \\ \frac { 1 }{ 2 }

Answer 6

A fair die is thrown once
Total number of outcomes = 6
Prime numbers = 2, 3, 5 and even prime is 2
Probability of getting an even prime number = \\ \frac { 1 }{ 6 }  (a)

Question 7

 

A fair die is thrown once. The probability of getting a composite number is
(a) \\ \frac { 1 }{ 3 }
(b) \\ \frac { 1 }{ 6 }
(c) \\ \frac { 2 }{ 3 }
(d) 0

Answer 7

A fair die is thrown once
Total number of outcomes = 6
Composite numbers are 4, 6 = 2
Probability = \\ \frac { 2 }{ 6 }  = \\ \frac { 1 }{ 3 }  (a)

Question 8

If a fair dice is rolled once, then the probability of getting an even number or a number greater than 4 is
(a) \\ \frac { 1 }{ 2 }
(b) \\ \frac { 1 }{ 3 }
(c) \\ \frac { 5 }{ 6 }
(d) \\ \frac { 2 }{ 3 }

Answer 8

A fair dice is thrown once.
Total number of outcomes = 6
Even numbers or a number greater than 4 = 2, 4, 5, 6 = 4
Probability = \\ \frac { 4 }{ 6 }  = \\ \frac { 2 }{ 3 }  (d)

Question 9

Rashmi has a die whose six faces show the letters as given below :
\boxed { A } \quad \boxed { B } \quad \boxed { C } \quad \boxed { D } \quad \boxed { A } \quad \boxed { C }
If she throws the die once, then the probability of getting C is
(a) \\ \frac { 1 }{ 3 }
(b) \\ \frac { 1 }{ 4 }
(c) \\ \frac { 1 }{ 5 }
(d) \\ \frac { 1 }{ 6 }

Answer 9

A die having 6 faces bearing letters A, B, C, D, A, C
Total number of outcomes = 4
Probability of getting C = \\ \frac { 2 }{ 6 }  = \\ \frac { 1 }{ 3 }  (a)

Question 10

 

If a letter is chosen at random from the letters of English alphabet, then the probability that it is a letter of the word ‘DELHI’ is
(a) \\ \frac { 1 }{ 5 }
(b) \\ \frac { 1 }{ 26 }
(c) \\ \frac { 5 }{ 26 }
(d) \\ \frac { 21 }{ 26 }

Answer 10

Total number of English alphabets = 26
Letter of Delhi = D, E, L, H, I. = 5
Probability = \\ \frac { 5 }{ 26 }  (c)

Question 11

A card is drawn from a well-shuffled pack of 52 playing cards. The event E is that the card drawn is not a face card. The number of outcomes favourable to the event E is
(a) 51
(b) 40
(c) 36
(d) 12

Answer 11

Number of playing cards = 52
Probability of a card which is not a face card = (52 – 12) = 40
Number of possible events = 40 (b)

Question 12

A card is drawn from a deck of 52 cards. The event E is that card is not an ace of hearts. The number of outcomes favourable to E is
(a) 4
(b) 13
(c) 48
(d) 51

Answer 12

Total number of cards = 52
Balance 52 – 1 = 51
Number of possible events = 51 (d)

Question 13

If one card is drawn from a well-shuffled pack of 52 cards, the probability of getting an ace is
(a) \\ \frac { 1 }{ 52 }
(b) \\ \frac { 4 }{ 13 }
(c) \\ \frac { 2 }{ 13 }
(d) \\ \frac { 1 }{ 13 }

Answer 13

Total number of cards = 52
Number of aces = 4
Probability of card being an ace = \\ \frac { 4 }{ 52 }  = \\ \frac { 1 }{ 13 }  (d)

Question 14

A card is selected at random from a well- shuffled deck of 52 cards. The probability of its being a face card is
(a) \\ \frac { 3 }{ 13 }
(b) \\ \frac { 4 }{ 13 }
(c) \\ \frac { 6 }{ 13 }
(d) \\ \frac { 9 }{ 13 }

Answer 14

Total number of cards = 52
No. of face cards = 3 × 4 = 12
.’. Probability of face card = \\ \frac { 12 }{ 52 }  = \\ \frac { 3 }{ 13 }  (a)

Question 15

A card is selected at random from a pack of 52 cards. The probability of its being a red face card is
(a) \\ \frac { 3 }{ 26 }
(b) \\ \frac { 3 }{ 13 }
(c) \\ \frac { 2 }{ 13 }
(d) \\ \frac { 1 }{ 2 }

Answer 15

Total number of card = 52
No. of red face card = 3 × 2 = 6
.’. Probability = \\ \frac { 6 }{ 52 }  = \\ \frac { 3 }{ 26 }  (a)

Question 16

If a card is drawn from a well-shuffled pack of 52 playing cards, then the probability of this card being a king or a jack is
(a) \\ \frac { 1 }{ 26 }
(b) \\ \frac { 1 }{ 13 }
(c) \\ \frac { 2 }{ 13 }
(d) \\ \frac { 4 }{ 13 }

Answer 16

Total number of cards 52
Number of a king or a jack = 4 + 4 = 8
.’. Probability = \\ \frac { 8 }{ 52 }  = \\ \frac { 2 }{ 13 }  (c)

Question 17

The probability that a non-leap year selected at random has 53 Sundays is.
(a) \\ \frac { 1 }{ 365 }
(b) \\ \frac { 2 }{ 365 }
(c) \\ \frac { 2 }{ 7 }
(d) \\ \frac { 1 }{ 7 }

Answer 17

Number of a non-leap year 365
Number of Sundays = 53
In a leap year, there are 52 weeks or 364 days
One days is left
Now we have to find the probability of a Sunday out of remaining 1 day
∴ Probability = \\ \frac { 1 }{ 7 }  (d)

Question 18

A bag contains 3 red balk, 5 white balls and 7 black balls. The probability that a ball drawn from the bag at random will be neither red nor black is
(a) \\ \frac { 1 }{ 5 }
(b) \\ \frac { 1 }{ 3 }
(c) \\ \frac { 7 }{ 15 }
(d) \\ \frac { 8 }{ 1 }

Answer 18

In a bag, there are
3 red balls + 5 white balls + 7 black balls
Total number of balls = 15
One ball is drawn at random which is neither
red not black
Number of outcomes = 5
Probability = \\ \frac { 5 }{ 15 }  = \\ \frac { 1 }{ 3 }  (b)

Question 19

A bag contains 4 red balls and 5 green balls. One ball is drawn at random from the bag. The probability of getting either a red ball or a green ball is
(a) \\ \frac { 4 }{ 9 }
(b) \\ \frac { 5 }{ 9 }
(c) 0
(d) 1

Answer 19

In a bag, there are
4 red balls + 5 green balls
Total 4 + 5 = 9
One ball is drawn at random
Probability of either a red or a green ball = \\ \frac { 9 }{ 9 }  = 1 (d)

Question 20

A bag contains 5 red, 4 white and 3 black balls. If a. ball is drawn from the bag at random, then the probability of the ball being not black is
(a) \\ \frac { 5 }{ 12 }
(b) \\ \frac { 1 }{ 3 }
(c) \\ \frac { 3 }{ 4 }
(d) \\ \frac { 1 }{ 4 }

Answer 20

In a bag, there are
5 red + 4 white + 3 black balls = 12
One ball is drawn at random
Probability of a ball not black = \\ \frac { 5+4 }{ 12 }  = \\ \frac { 9 }{ 12 }  = \\ \frac { 3 }{ 4 }  (c)

Question 21

One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number which is a multiple of 5 is
(a) \\ \frac { 1 }{ 5 }
(b) \\ \frac { 3 }{ 5 }
(c) \\ \frac { 4 }{ 5 }
(d) \\ \frac { 1 }{ 3 }

Answer 21

There are t to 40 = 40 tickets in a bag
No. of tickets which is multiple of 5 = 8
(5, 10, 15, 20, 25, 30, 35, 40)
Probability = \\ \frac { 8 }{ 40 }  = \\ \frac { 1 }{ 5 }  (a)

Question 22

If a number is randomly chosen from the numbers 1,2,3,4, …, 25, then the probability of the number to be prime is
(a) \\ \frac { 7 }{ 25 }
(b) \\ \frac { 9 }{ 25 }
(c) \\ \frac { 11 }{ 25 }
(d) \\ \frac { 13 }{ 25 }

Answer 22

There are 25 number bearing numbers 1, 2, 3,…,25
Prime numbers are 2, 3, 5, 7, 11, 13, 17 19, 23 = 9
Probability being a prime number = \\ \frac { 9 }{ 25 }  (b)

Question 23

A box contains 90 cards numbered 1 to 90. If one card is drawn from the box at random, then the probability that the number on the card is a perfect square is
(a) \\ \frac { 1 }{ 10 }
(b) \\ \frac { 9 }{ 100 }
(c) \\ \frac { 1 }{ 9 }
(d) \\ \frac { 1 }{ 100 }

Answer 23

In a box, there are
90 cards bearing numbers 1 to 90
Perfect squares are 1, 4, 9, 16, 25, 36, 49, 64, 81 = 9
Probability of being a perfect square = \\ \frac { 9 }{ 90 }  = \\ \frac { 1 }{ 10 }  (a)

Question 24

If a (fair) coin is tossed twice, then the probability of getting two heads is
(a) \\ \frac { 1 }{ 4 }
(b) \\ \frac { 1 }{ 2 }
(c) \\ \frac { 3 }{ 4 }
(d) 0

Answer 24

A coin is tossed twice
Number of outcomes = 2 x 2 = 4
Probability of getting two heads (HH = 1) = \\ \frac { 1 }{ 4 }  (a)

Question 25

If two coins are tossed simultaneously, then the probability of getting atleast one head is
(a) \\ \frac { 1 }{ 4 }
(b) \\ \frac { 1 }{ 2 }
(c) \\ \frac { 3 }{ 4 }
(d) 1

Answer 25

Two coins are tossed
Total outcomes = 2 × 2 = 4
Probability of getting atleast one head (HT,TH,H,H) = \\ \frac { 3 }{ 4 }  (c)

Question 26

Lakshmi tosses two coins simultaneously. The probability that she gets almost one head
(a) 1
(b) \\ \frac { 3 }{ 4 }
(c) \\ \frac { 1 }{ 2 }
(d) \\ \frac { 1 }{ 7 }

Answer 26

Two coins are tossed
Total number of outcomes = 2 × 2 = 4
Probability of getting atleast one head = (HT, TH, RH = 3) = \\ \frac { 3 }{ 4 }  (b)

Question 27

The probability of getting a bad egg in a lot of 400 eggs is 0.035. The number of bad eggs in the lot is
(a) 7
(b) 14
(c) 21
(d) 28

Answer 27

Total number of eggs 400
Probability of getting a bad egg = 0.035
Number of bad eggs = 0.035 of 400 = 400 \times \frac { 35 }{ 1000 }  = 14 (b)

Question 28

.
A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, how many tickets she has bought?
(a) 40
(b) 240
(c) 480
(d) 750

Answer 28

For a girl,
Winning a first prize = 0.08
Number of total tickets = 6000
Number of tickets she bought = 0.08 of 6000 = 6000 \times \frac { 8 }{ 100 } = 480 (c)


Probability Chapter 22 ML Aggarwal Solution Chapter-Test

Question 1

A game consists of spinning an arrow which comes to rest at one of the regions 1, 2 or 3 (shown in the given figure). Are the outcomes 1, 2 and 3 equally likely to occur? Give reasons.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Chapter Test 1

Answer 1

In a game,
No, the outcomes are not equally likely.
Outcome 3 is more likely to occur than the outcomes of 1 and 2.

Question 2

In a single throw of a die, find the probability of getting
(i) a number greater than 5
(ii) an odd prime number
(iii) a number which is multiple of 3 or 4.

Answer 2

In a single throw of a die
Number of total outcomes = 6 (1, 2, 3, 4, 5, 6)
(i) Numbers greater than 5 = 6 i.e., one number
Probability = \\ \frac { 1 }{ 6 }
(ii) An odd prime number 2 i.e., one number
Probability = \\ \frac { 1 }{ 6 }
(iii) A number which is a multiple of 3 or 4 which are 3, 6, 4 = 3 numbers
Probability = \\ \frac { 3 }{ 6 }  = \\ \frac { 1 }{ 2 }

Question 3

A lot consists of 144 ball pens of which 20 are defective and the others are good. Rohana will buy a pen if it is good, but will not buy it if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that :
(i) She will buy it?
(ii) She will not buy it?

Answer 3

In a lot, there are 144 ball pens in which defective ball pens are = 20
and good ball pens are = 144 – 20 = 124
Rohana buys a pen which is good only.
(i) Now the number of possible outcomes = 144
and the number of favourable outcomes = 124
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Chapter Test 3
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Chapter Test 3.1

Question 4

A lot consists of 48 mobile phones of which 42 are good, 3 have only minor defects and 3 have major defects. Varnika will buy a phone if it is good but the trader will only buy a mobile if it has no major defect. One phone is selected at random from the lot. What is the probability that it is
(i) acceptable to Varnika?
(ii) acceptable to the trader?

Answer 4

Number of total mobiles = 48,Number of good mobiles = 42,Number having minor defect = 3
and Number having major defect = 3
(i) Acceptable to Varnika = 42
Probability = \\ \frac { 42 }{ 48 }  = \\ \frac { 7 }{ 8 }
(ii) Acceptable to trader = 42 + 3 = 45
Probability = \\ \frac { 45 }{ 48 }  = \\ \frac { 15 }{ 16 }

Question 5

A bag contains 6 red, 5 black and 4 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is
(i) white
(ii) red
(iii) not black
(iv) red or white.

Answer 5

Total number of balls = 6 + 5 + 4 = 15, Total Number of red balls = 6
and Number of black balls = 5
Number of white balls = 4
(i) Probability of a white ball will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 4 }{ 15 }
(ii) Probability of red ball will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 6 }{ 15 }  = \\ \frac { 2 }{ 5 }
(iii) Probability of not black ball will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 15-5 }{ 15 }
\\ \frac { 10 }{ 15 }
\\ \frac { 2 }{ 3 }
(iv) Probability of red or white ball will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 6+4 }{ 15 }
\\ \frac { 10 }{ 15 }
\\ \frac { 2 }{ 3 }

Question 6

A bag contains 5 red, 8 white and 7 black balls. A ball is drawn from the bag at random. Find the probability that the drawn ball is:
(i) red or white
(ii) not black
(iii) neither white nor black

Answer 6

Total number of balls in a bag = 5 + 8 + 7 = 20
(i) Number of red or white balls = 5 + 8 = 13
Probability of red or white ball will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 13 }{ 20 }
(ii) Number of ball which are not black = 20 – 7 = 13
Probability of not black ball will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 13 }{ 20 }
(iii) Number of ball which are neither white nor black
= Number of ball which are only red = 5
Probability of neither white nor black ball will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }
\\ \frac { 5 }{ 20 }
\\ \frac { 1 }{ 4 }

Question 7

A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is :
(i) white or blue
(ii) red or black
(iii) not white
(iv) neither white nor black ?

Answer 7

Number of total balls = 5 + 7 + 4 + 2 = 18
Total Number of white balls = 5, number of red balls = 7,number of black balls = 4

and number of blue balls = 2.

(i) Number of white and blue balls = 5 + 2 = 7

Probability of white or blue balls will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 7 }{ 18 }

(ii) Number of red and black balls = 7 + 4 = 11

Probability of red or black balls will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 11 }{ 18 }

(iii) Number of ball which are not white = 7 + 4 + 2 = 13

Probability of not white balls will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 13 }{ 18 }

(iv) Number of balls which are neither white nor black = 18 – (5 + 4) = 18 – 9 = 9

Probability of ball which is neither white nor black will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 9 }{ 18 }  = \\ \frac { 1 }{ 2 }

Question 8

A box contains 20 balls bearing numbers 1, 2, 3, 4,……, 20. A ball is drawn at random from the box. What is the probability that the number on the ball is
(i) an odd number
(ii) divisible by 2 or 3
(iii) prime number
(iv) not divisible by 10?

Answer 8

In a box, there are 20 balls containing 1 to 20 number
Number of possible outcomes = 20

(i) Numbers which are odd will be,

1, 3, 5, 7, 9, 11, 13, 15, 17, 19 = 10 balls.
Probability of odd ball will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 10 }{ 20 }  = \\ \frac { 1 }{ 2 }

(ii) Numbers which are divisible by 2 or 3 will be

2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20 = 13 balls
Probability of ball which is divisible by 2 or 3 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 13 }{ 20 }

(iii) Prime numbers will be 2, 3, 5, 7, 11, 13, 17, 19 = 8

Probability of prime number will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 8 }{ 20 }  = \\ \frac { 2 }{ 5 }

(iv) Numbers not divisible by 10 will be

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19 = 18
Probability of prime number will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 18 }{ 20 }  = \\ \frac { 9 }{ 10 }

Question 9

Find the probability that a number selected at random from the numbers 1, 2, 3,……35 is a
(i) prime number
(ii) multiple of 7
(iii) multiple of 3 or 5.

Answer 9

Numbers are 1, 2, 3, 4, 5,…..30, 31, 32, 33, 34, 35
Total = 35

(i) Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31

which are 11
Probability of prime number will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 11 }{ 35 }

(ii) Multiple of 7 are 7, 14, 21, 28, 35 which are 5

Probability of multiple of 7 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 5 }{ 35 }  = \\ \frac { 1 }{ 7 }

(iii) Multiple of 3 or 5 are 3, 5, 6, 9, 10, 12 ,15, 18, 20, 21, 24, 25, 27, 30, 33, 35.

Which are 16 in numbers
Probability of multiple of 3 or 5 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 16 }{ 35 }

Question 10

Cards marked with numbers 13, 14, 15,…..60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card is
(i) divisible by 5
(ii) a number which is a perfect square.

Answer 10

Number of cards which are marked with numbers
13, 14, 15, 16, 17,….to 59, 60 are = 48

(i) Numbers which are divisible by 5 will be

15, 20, 25, 30, 35, 40, 45, 50, 55, 60 = 10
Probability of number divisible by 5 will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 10 }{ 48 }  = \\ \frac { 5 }{ 24 }

(ii) Numbers which is a perfect square are 16, 25, 36, 49 which are 4 in numbers.

Probability of number which is a perfect square will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 4 }{ 48 }  = \\ \frac { 1 }{ 12 }

Question 11

The box has cards numbered 14 to 99. Cards are mixed thoroughly and a card is drawn at random from the box. Find the probability that the card drawn from the box has
(i) an odd number
(ii) a perfect square number.

Answer 11

Cards in a box are from 14 to 99 = 86
No. of total cards = 86
One card is drawn at random
Cards bearing odd numbers are 15, 17, 19, 21, …, 97, 99
Which are 43

(i) P(E) = \frac { Number\quad of\quad actual\quad events }{ Number\quad of\quad total\quad events }

= \\ \frac { 43 }{ 86 }
\\ \frac { 1 }{ 2 }

(ii) Cards bearing number which are a perfect square

= 16, 25, 36, 49, 64, 81
Which are 6
P(E) = \frac { Number\quad of\quad actual\quad events }{ Number\quad of\quad total\quad events }
\\ \frac { 6 }{ 86 }
\\ \frac { 3 }{ 43 }

Question 12

A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is four times that of a red ball, find the number of balls in the bags.

Answer 12

Number of red balls = 5
and let number of blue balls = x
Total balls in the bag = 5 + x
and that of red balls = \\ \frac { 5 }{ 5+x }
According to the condition,
\frac { x }{ 5+x } =4\times \frac { 5 }{ 5+x } =>\frac { x }{ 5+x } =\frac { 20 }{ 5+x }
x ≠ – 5
x = 20
Hence, number of blue balls = 20
and number of balls in the bag = 20 + 5 = 25

Question 13

A bag contains 18 balls out of which x balls are white.
(i) If one ball is drawn at random from the bag, what is the probability that it is white ball?

(ii) If 2 more white balls are put in the bag, the probability of drawing a white ball will be \\ \frac { 9 }{ 8 }  times that of probability of white ball coming in part (i). Find the value of x.

Answer 13

Total numbers of balls in a bag = 18
No. of white balls = x
(i) One ball is drawn a random
Probability of being a white ball = \\ \frac { x }{ 18 }
(ii) If 2 more white balls an put, then number of white balls = x + 2
and probability is \\ \frac { 9 }{ 8 }  times
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Chapter Test 13

Question 14

A card is drawn from a well-shuffled pack of 52 cards. Find the probability that the card drawn is :
(i) a red face card
(ii) neither a club nor a spade
(iii) neither an ace nor a king of red colour
(iv) neither a red card nor a queen
(v) neither a red card nor a black king.

Answer 14

Number of cards in a pack of well-shuffled cards = 52
(i) Number of a red face card = 3 + 3 = 6
Probability of red face card will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 6 }{ 52 }  = \\ \frac { 3 }{ 26 }
(ii) Number of cards which is neither a club nor a spade = 52 – 26 = 26
Probability of card which’ is neither a club nor a spade will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 26 }{ 52 }  = \\ \frac { 1 }{ 2 }

(iii) Number of cards which is neither an ace nor a king of red colour

= 52 – (4 + 2) = 52 – 6 = 46
Probability of card which is neither ace nor a king of red colour will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 46 }{ 52 }  = \\ \frac { 23 }{ 26 }
(iv) Number of cards which are neither a red card nor a queen are
= 52 – (26 + 2) = 52 – 28 = 24
Probability of card which is neither red nor a queen will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 24 }{ 52 }  = \\ \frac { 6 }{ 13 }
(v) Number of cards which are neither red card nor a black king
= 52 – (26 + 2) = 52 – 28 = 24
Probability of cards which is neither red nor a black king will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 24 }{ 52 }  = \\ \frac { 6 }{ 13 }

Question 15

From pack of 52 playing cards, blackjacks, black kings and black aces are removed and then the remaining pack is well-shuffled. A card is drawn at random from the remaining pack. Find the probability of getting
(i) a red card
(ii) a face card
(iii) a diamond or a club
(iv) a queen or a spade.

Answer 15

Total number of cards = 52
Black jacks, black kings and black aces are removed
Now number of cards = 52 – (2 + 2 + 2) = 52 – 6 = 46
One card is drawn
(i) No. of red cards = 13 + 13 = 26
∴Probability = \\ \frac { 26 }{ 46 }  = \\ \frac { 13 }{ 23 }

(ii) Face cards = 4 queens, 2 red jacks, 2 kings = 8

∴Probability = \\ \frac { 8 }{ 46 }  = \\ \frac { 4 }{ 23 }
(iii) a diamond on a club = 13 + 10 = 23
∴Probability = \\ \frac { 23 }{ 46 }  = \\ \frac { 1 }{ 2 }
(iv) A queen or a spade = 4 + 10 = 14
∴Probability = \\ \frac { 14 }{ 46 }  = \\ \frac { 7 }{ 23 }

Question 16

Two different dice are thrown simultaneously. Find the probability of getting:
(i) sum 7
(ii) sum ≤ 3
(iii) sum ≤ 10

Answer 16

(i) Numbers whose sum is 7 will be (1, 6), (2, 5), (4, 3), (5, 2), (6, 1), (3, 4) = 6
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 6 }{ 36 }  = \\ \frac { 1 }{ 6 }
(ii) Sum ≤ 3
Then numbers can be (1, 2), (2, 1), (1, 1) which are 3 in numbers
∴Probability will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 3 }{ 36 }  = \\ \frac { 1 }{ 12 }
(iii) Sum ≤ 10
The numbers can be,
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, .6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5),
(6, 1), (6, 2), (6, 3), (6, 4) = 33
Probability will be
P(E) = \frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }  = \\ \frac { 33 }{ 36 }  = \\ \frac { 11 }{ 12 }

Question 17

Two dice are thrown together. Find the probability that the product of the numbers on the top of two dice is
(i) 6
(ii) 12
(iii) 7

Answer 17

Two dice are thrown together
Total number of events = 6 × 6 = 36
(i) Product 6 = (1, 6), (2, 3), (3, 2). (6, 1) = 4
Probability = \\ \frac { 4 }{ 36 }  = \\ \frac { 1 }{ 9 }
(ii) Product 12 = (2, 6), (3, 4), (4, 3), (6, 2) = 4
Probability = \\ \frac { 4 }{ 36 }  = \\ \frac { 1 }{ 9 }
(iii) Product 7 = 0 (no outcomes)
Probability = \\ \frac { 0 }{ 36 }  = 0

 

–: End of Probability ML Aggarwal Solutions  :–


 

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