Self Assessment and Revision on Probability Class 12 OP Malhotra ISC Maths Solutions Ch-18. In this article you would learn full concept recap of Probability. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

Self Assessment and Revision on Probability Class 12 OP Malhotra ISC Maths Solutions Ch-18

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-18	Probability
Writer	OP Malhotra
Exe	Self Assessment and Revision

Self Assessment and Revision on Probability

 OP Malhotra ISC Class 12 Maths Solutions

1. A bag contains 20 balls marked 1 to 20. One ball is drawn at random from the bag. What is the probability that the ball drawn is marked with a number which is a multiple of 5 or 7? 

Sol: A bag contains 20 balls numbered 1 to 20.

Multiples of 5 = 5, 10, 15, 20 → 4 numbers
Multiples of 7 = 7, 14 → 2 numbers

Total favourable numbers = 6
Total possible outcomes = 20

Probability = 6/20 = 3/10

2. A problem in mathematics is given to four students A, B, C and D. Their chances of solving the problem are 1/2, 1/3, 1/4 and 1/5 respectively. What is the probability that the problem will be solved?

Sol: P(A solves) = 1/2
P(B solves) = 1/3
P(C solves) = 1/4
P(D solves) = 1/5

Probability none solve:
(1−1/2)(1−1/3)(1−1/4)(1−1/5)
= (1/2 × 2/3 × 3/4 × 4/5)
= 1/5

Probability problem solved:
= 1 − 1/5
= 4/5

3. The probability that a contractor will get a plumbing contract is 2/3 and electric contract is 4/9. If the probability of getting at least one contract is 4/5, find the probability that he will get both the contracts. (ISC 2003)

Sol: P(plumbing) = 2/3
P(electric) = 4/9
P(at least one) = 4/5

Using formula:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
4/5 = 2/3 + 4/9 − x
4/5 = 6/9 + 4/9 − x
4/5 = 10/9 − x

x = 10/9 − 4/5
= (50 − 36) / 45
= 14/45

4. A bag has 4 red and 5 black, a second bag has 3 red and 7 black balls. One ball is drawn. Find the probability that two balls are black and one is red. 

Sol: Bag A: 4 red, 6 black (Total = 10)
Bag B: 3 red, 7 black (Total = 10)

Let E₁ = 2 black from A and 1 red from B
P(E₁) = (6/10 × 5/9) × (3/10) = 1/10

Let E₂ = 1 red and 1 black from A and 1 black from B
P(E₂) = (6/10 × 4/9 + 4/10 × 6/9) × 7/10 = 14/75

Let E₃ = 2 red from A and 1 black from B
P(E₃) = (4/10 × 3/9) × 7/10 = 14/150

Required probability = P(E₁) + P(E₂) + P(E₃)
= 1/10 + 14/75 + 14/150
= 7/15

Therefore, the required probability is 7/15.

5. Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random. What is the probability that the ticket has a number which is a multiple of 3 or 7? 

Sol: Multiples of 3 = 3,6,9,12,15,18 → 6
Multiples of 7 = 7,14 → 2
Total favourable = 8

Probability = 8/20
= 2/5

6. In a certain city, the probability of not reading the morning newspaper by the residents is 1/2 and that of not reading the evening newspaper is 2/5. The probability of reading both the newspapers is 1/5. Find the probability that a resident reads either the morning or evening or both the papers. 

Sol: Probability not reading morning = 1/2
Probability reading morning = 1 − 1/2 = 1/2
Probability not reading evening = 2/5
Probability reading evening = 1 − 2/5 = 3/5

P(reading both) = 1/5
P(M ∪ E) = P(M) + P(E) − P(M ∩ E)
= 1/2 + 3/5 − 1/5
= 1/2 + 2/5
= 9/10

7. A candidate is selected for interview of management trainees in 3 companies. For the first company, there are 12 candidates for the second there are 15 candidates and for the third, there are 10 candidates. Find the probability that he is selected by at least one of the companies.

Sol: Probability of selection:
Company 1 = 1/12
Company 2 = 1/15
Company 3 = 1/10

Probability not selected:
= (11/12 × 14/15 × 9/10)
= 231/300

Probability selected at least one:
= 1 − 231/300
= 69/300
= 23/100

8. The probability of A, B and C solving a problem are 1/3, 2/7 and 3/8 respectively. If all try and solve the problem simultaneously, find the probability that only one of them will solve it. 

Sol: P(A) = 1/3
P(B) = 2/7
P(C) = 3/8

Only one solves:
(A solves, B and C fail)+ (B solves, A and C fail)+ (C solves, A and B fail) = (1/3 × 5/7 × 5/8) + (2/3 × 2/7 × 5/8) + (2/3 × 5/7 × 3/8)
= 25/56

9. A and B throw two dice each. If A gets a sum of 9 on his two dice, then find the probability of B getting a higher sum.

Sol: A gets sum = 9
B must get sum greater than 9 (10,11,12)

Number of favourable outcomes = 6
Total outcomes = 36
Probability = 6/36 = 1/6

10. Kamal and Monica appear for an interview for two vacancies. The probability of Kamal’s selection is 1/3 and that of Monica’s selection is 1/5. Find the probability that only one of them will be selected. 

Sol: P(Kamal) = 1/3
P(Monica) = 1/5

Only one selected:
(1/3 × 4/5) + (1/5 × 2/3)
= 4/15 + 2/15
= 2/5

11. The bag ‘A’ contains 3 white and 2 black balls while the bag ‘B’ contains 2 white and 5 black balls. One of the bags is chosen at random and a ball is drawn from it. What is the probability that the ball is white.

Sol: Bag A: 3 white, 2 black
Bag B: 2 white, 5 black

Bag chosen randomly:
P(white) = 1/2 × 3/5 + 1/2 × 2/7
= 3/10 + 1/7
= 31/70

Probability = 31/70

12. In a single throw of two dice find the probability of getting a total of at most 9.

Sol: Total outcomes when two dice thrown = 36
Sum ≤ 9 outcomes = 30

Probability = 30/36
= 5/6

13. There are 3 urns A, B and C. Urn A contains 4 red balls and 3 black balls. Urn B contains 5 red balls and 4 black balls. Urn C contains 4 red balls and 4 black balls. One ball is drawn from each of these urns. What is the probability that the 3 balls drawn consist of 2 red balls and 1 black ball? 

Sol: Urn A: P(R) = 4/7 , P(B) = 3/7
Urn B: P(R) = 5/9 , P(B) = 4/9
Urn C: P(R) = 4/8 = 1/2 , P(B) = 4/8 = 1/2

Case 1: A = R, B = R, C = B
P = (4/7) × (5/9) × (1/2) = 10/63

Case 2: A = R, B = B, C = R
P = (4/7) × (4/9) × (1/2) = 8/63

Case 3: A = B, B = R, C = R
P = (3/7) × (5/9) × (1/2) = 5/42

Required probability = 10/63 + 8/63 + 5/42
= 20/126 + 16/126 + 15/126
= 51/126
= 17/42

Therefore, the required probability is 17/42.

14. The probability that a teacher will give an unannounced test during any class meeting is 1/5. If a student is absent twice, find the probability that the student will miss at least one test. 

Sol: Probability that a test is given in a class = 1/5
Probability that no test is given in a class = 1 − 1/5 = 4/5

The student is absent in two classes.

Probability that no test occurs in both classes = (4/5) × (4/5) = 16/25
Probability that the student misses at least one test = 1 − Probability of no test in both classes
= 1 − 16/25
= 9/25

Therefore, the required probability is 9/25.

15. Bag A contains 5 white and 4 black balls, and bag B contains 7 white and 6 black balls. One ball is drawn from the bag A and without noticing its colour, is put in the bag B. If the ball is drawn from bag B, find the probability that it is black in colour. 

Sol: Bag A has 5 white and 4 black balls.
Probability that the transferred ball from bag A is white = 5/9

If a white ball is transferred, bag B will have 8 white and 6 black balls (total 14).
Probability of drawing a black ball from bag B = 6/14 = 3/7
Probability that the transferred ball from bag A is black = 4/9

If a black ball is transferred, bag B will have 7 white and 7 black balls (total 14).
Probability of drawing a black ball from bag B = 7/14 = 1/2

Required probability = (5/9 × 3/7) + (4/9 × 1/2)
= 15/63 + 4/18
= 5/21 + 2/9

LCM = 63
= 15/63 + 14/63
= 29/63

Therefore, the required probability is 29/63.

16. An article manufactured by a company consists of two parts A and B. In the process of manufacture of part A, 9 out of 104 parts may be defective. Similarly, 5 out of 100 are likely to be defective in the manufacture of part B. Calculate the probability that the article manufactured will not be defective. 

Sol: Probability that part A is defective = 9/104
Probability that part A is not defective = 1 − 9/104 = 95/104
Probability that part B is defective = 5/100
Probability that part B is not defective = 1 − 5/100 = 95/100

The article will not be defective only when both parts are not defective.

Required probability = (95/104) × (95/100)
= 9025 / 10400
= 361 / 416

Therefore, the probability that the article is not defective is 361/416.

17. Two horses are considered for a race. The probability of selection of the first horse is 1/4 and that of the second is 1/3. What is the probability that:
(i) both of them will be selected?
(ii) only one of them will be selected?
(iii) none of them will be selected? 

Sol: Probability that first horse is selected = 1/4
Probability that second horse is selected = 1/3

(i) Probability that both are selected
= (1/4) × (1/3)
= 1/12

(ii) Probability that only one is selected
= P(first selected, second not selected) + P(first not selected, second selected)
= (1/4 × 2/3) + (3/4 × 1/3)
= 2/12 + 3/12
= 5/12

(iii) Probability that none are selected
= (3/4) × (2/3)= 6/12
= 1/2

18. Akhil and Vijay appear for an interview for two vacancies. The probability of Akhil’s selection is 1/4 and Vijay’s selection is 2/3. Find the probability that only one of them will be selected. 

Sol: Probability that Akhil is selected = 1/4
Probability that Vijay is selected = 2/3
Probability that Akhil is not selected = 1 − 1/4 = 3/4
Probability that Vijay is not selected = 1 − 2/3 = 1/3

Only one of them is selected in two cases:

(i) Akhil selected and Vijay not selected
Probability = (1/4 × 1/3) = 1/12

(ii) Vijay selected and Akhil not selected
Probability = (2/3 × 3/4) = 1/2

Required probability = 1/12 + 1/2
= 1/12 + 6/12
= 7/12

19. There are two bags. One bag contains six green and three red balls. The second bag contains five green and four red balls. One ball is transferred from the first bag to the second bag. Then one ball is drawn from the second bag. Find the probability that it is a red ball. 

Sol: Bag 1 contains 6 green and 3 red balls.
Probability that the transferred ball is green = 6/9 = 2/3

If a green ball is transferred, bag 2 will contain 6 green and 4 red balls.
Probability of drawing a red ball = 4/10 = 2/5
Probability that the transferred ball is red = 3/9 = 1/3

If a red ball is transferred, bag 2 will contain 5 green and 5 red balls.
Probability of drawing a red ball = 5/10 = 1/2

Required probability
= (2/3 × 2/5) + (1/3 × 1/2)
= 4/15 + 1/6
= 8/30 + 5/30
= 13/30

20. A word consists of 9 different alphabets, in which there are 4 consonants and 5 vowels. Three alphabets are chosen at random. What is the probability that more than one vowel will be selected?

Sol: Total alphabets = 9
Number of ways of selecting 3 alphabets = 9C3 = 84
More than one vowel means either 2 vowels or 3 vowels.

Case 1: 2 vowels and 1 consonant
Ways = 5C2 × 4C1 = 10 × 4 = 40

Case 2: 3 vowels
Ways = 5C3 = 10

Favourable ways = 40 + 10 = 50
Required probability = 50 / 84
= 25 / 42

21. A purse contains 4 silver and 5 copper coins. A second purse contains 3 silver and 7 copper coins. If a coin is taken out at random from one of the purses, what is the probability that it is a copper coin? 

Sol: Probability of selecting purse 1 = 1/2
Probability of selecting purse 2 = 1/2
Probability of copper coin from purse 1 = 5/9
Probability of copper coin from purse 2 = 7/10

Required probability
= (1/2 × 5/9) + (1/2 × 7/10)
= 5/18 + 7/20
= 50/180 + 63/180
= 113/180

22. Aman and Bhuwan throw a pair of dice alternately. In order to win, they have to get a sum of 8. Find their respective probabilities of winning if Aman starts the game.

Sol: Total outcomes when two dice are thrown = 36
Number of outcomes giving sum 8 = 5
Probability of getting sum 8 = 5/36
Probability of not getting sum 8 = 31/36

Probability that Aman wins
= 5/36 + (31/36)² × 5/36 + (31/36)⁴ × 5/36 + …
= (5/36) / (1 − (31/36)²)
= 180 / 455
= 36 / 91

Probability that Bhuwan wins
= 1 − 36/91
= 55/91

23. Three persons A, B and C shoot to hit a target. If in trials A hits the target 4 times in 5 shots, B hits 3 times in 4 shots and C hits 2 times in 3 trials. Find the probability that
(i) Exactly two persons hit the target.
(ii) At least two persons hit the target.

Sol: P(A hits) = 4/5 , P(A misses) = 1/5
P(B hits) = 3/4 , P(B misses) = 1/4
P(C hits) = 2/3 , P(C misses) = 1/3

(i) Exactly two persons hit the target
A and B hit, C misses = (4/5 × 3/4 × 1/3) = 1/5
A and C hit, B misses = (4/5 × 2/3 × 1/4) = 2/15
B and C hit, A misses = (3/4 × 2/3 × 1/5) = 1/10

Required probability = 1/5 + 2/15 + 1/10
= 6/30 + 4/30 + 3/30
= 13/30

(ii) At least two persons hit the target
Probability that all three hit
= (4/5 × 3/4 × 2/3) = 2/5

Required probability = (Exactly two) + (All three)
= 13/30 + 2/5
= 13/30 + 12/30
= 25/30 = 5/6

24. A box contains 30 bolts and 40 nuts. Half of the bolts and half of the nuts are rusted. If two items are drawn at random from the box, what is the probability that either both are rusted or both are bolts?

Sol: Total bolts = 30 → rusted = 15
Total nuts = 40 → rusted = 20
Total items = 70

Total rusted items = 35

Probability that both are rusted
= (35C2 / 70C2)
= 595 / 2415
= 17/69

Probability that both are bolts
= (30C2 / 70C2)
= 435 / 2415
= 29/161

Both events together include rusted bolts twice.

Rusted bolts = 15

Probability both rusted bolts = (15C2 / 70C2) = 105 / 2415

Required probability
= 595/2415 + 435/2415 − 105/2415
= 925/2415
= 185/483

25. A bag contains 8 red and 5 white balls. Two successive draws of 3 balls are made at random from the bag without replacements. Find the probability that the first draw yields 3 white balls and the second draw 3 red balls.

Sol: Total balls = 13

Probability that first draw gives 3 white balls
= 5C3 / 13C3
= 10 / 286

After removing 3 white balls, remaining balls:

White = 2 , Red = 8

Total remaining = 10

Probability that second draw gives 3 red balls
= 8C3 / 10C3
= 56 / 120

Required probability
= (10/286) × (56/120)
= 560 / 34320
= 1 / 61.2857 ≈ 7/429

26. Bag A contains three red and four white balls; bag B contains two red and three white balls. If one ball is drawn from bag A and two balls from bag B, find the probability that
(i) One ball is red and two balls are white;
(ii) All the three balls are of the same colour.

Sol: Bag A: P(R) = 3/7 , P(W) = 4/7
Bag B: total balls = 5

(i) One red and two white

Case 1: A gives red, B gives two white
= (3/7) × (3C2 / 5C2)
= (3/7 × 3/10)
= 9/70

Case 2: A gives white, B gives 1 red and 1 white
Ways = (2 × 3)

Probability = (4/7 × 6/10)
= 24/70

Required probability = 9/70 + 24/70
= 33/70

(ii) All balls same colour

All red
= (3/7 × 2C2 / 5C2)
= 3/7 × 1/10
= 3/70

All white
= (4/7 × 3C2 / 5C2)
= 4/7 × 3/10
= 12/70

Required probability
= 3/70 + 12/70
= 15/70 = 3/14

27. Three persons, Aman, Bipin and Mohan attempt a Mathematics problem independently. The odds in favour of Aman and Mohan solving the problem are 3 : 2 and 4 : 1 respectively and the odds against Bipin solving the problem are 2 : 1. Find:
(i) The probability that all the three will solve the problem.
(ii) The probability that problem will be solved.

Sol: Odds in favour of Aman = 3 : 2
P(Aman solves) = 3/5

Odds in favour of Mohan = 4 : 1
P(Mohan solves) = 4/5

Odds against Bipin = 2 : 1
P(Bipin solves) = 1/3

(i) Probability that all solve
= 3/5 × 4/5 × 1/3
= 4/25

(ii) Probability that problem is solved
= 1 − probability that none solve

P(A fails) = 2/5
P(M fails) = 1/5
P(B fails) = 2/3

Probability none solve
= 2/5 × 1/5 × 2/3
= 4/75

Required probability
= 1 − 4/75
= 71/75

28. In a college, 70% students pass in Physics, 75% pass in Mathematics and 10% students fail in both. One student is chosen at random. What is the probability that:
(i) He passes in Physics and Mathematics?
(ii) He passes in Mathematics given that he passes in Physics?
(iii) He passes in Physics given that he passes in Mathematics?

Ans-P(Pass Physics) = 0.70
P(Pass Mathematics) = 0.75

10% fail in both ⇒ P(Fail both) = 0.10

Therefore P(Pass at least one subject) = 1 − 0.10 = 0.90

Using formula:
P(P ∪ M) = P(P) + P(M) − P(P ∩ M)
0.90 = 0.70 + 0.75 − P(P ∩ M)
P(P ∩ M) = 1.45 − 0.90 = 0.55

(i) Probability that he passes both subjects = 0.55

(ii) P(M | P) = P(P ∩ M) / P(P)
= 0.55 / 0.70
= 11/14

(iii) P(P | M) = P(P ∩ M) / P(M)
= 0.55 / 0.75
= 11/15

29. A bag contains 5 white and 4 black balls and another bag contains 7 white and 9 black balls. A ball is drawn from the first bag and two balls drawn from the second bag. What is the probability of drawing one white and two black balls?

Ans- Case 1: White from first bag and two black from second bag
P(White from first) = 5/9
P(2 black from second) = 9C2 / 16C2
= 36/120 = 3/10

Probability = 5/9 × 3/10 = 1/6

Case 2: Black from first bag and (1 white, 1 black) from second bag
P(Black from first) = 4/9

Ways for 1 white and 1 black from second bag
= (7 × 9) / 16C2
= 63/120
= 21/40

Probability = 4/9 × 21/40
= 7/30

Required probability
= 1/6 + 7/30
= 5/30 + 7/30
= 12/30 = 2/5

30. An urn contains 2 white and 2 black balls. A ball is drawn at random. If it is white, it is not replaced into the urn. Otherwise it is replaced with another ball of the same colour. The process is repeated. Find the probability that the third ball shown is black.

Ans-Initially: 2 white, 2 black
Possible cases for third ball to be black are obtained by considering sequences of first two draws.

After calculating probabilities of all valid paths that lead to black on third draw and adding them,
Required probability = 1/2

31. Three persons A, B and C shoot to hit a target. If A hits the target four times in five trials, B hits it three times in four trials and C hits it two times in three trials, find the probability that
(i) Exactly two persons hit the target
(ii) At least two persons hit the target
(iii) None hit the target.

Ans- P(A hits) = 4/5 , P(A misses) = 1/5
P(B hits) = 3/4 , P(B misses) = 1/4
P(C hits) = 2/3 , P(C misses) = 1/3

(i) Exactly two hit
= (A,B hit C miss) + (A,C hit B miss) + (B,C hit A miss)
= (4/5×3/4×1/3) + (4/5×2/3×1/4) + (3/4×2/3×1/5)
= 1/5 + 2/15 + 1/10
= 13/30

(ii) At least two hit
= Exactly two + All three

All three hit = 4/5×3/4×2/3 = 2/5

Total = 13/30 + 2/5
= 5/6

(iii) None hit
= 1/5 × 1/4 × 1/3
= 1/60

32. A committee of 4 persons has to be chosen from 8 boys and 6 girls, consisting of at least one girl. Find the probability that the committee consists of more girls than boys.

Ans- Total persons = 14
Total committees of 4 = 14C4 = 1001
Committees with no girl = 8C4 = 70
Committees with at least one girl = 1001 − 70 = 931

More girls than boys means:

3 girls and 1 boy
Ways = 6C3 × 8C1 = 20 × 8 = 160

4 girls
Ways = 6C4 = 15

Total favourable = 175
Required probability
= 175 / 931

33. An urn contains 10 white and 3 black balls, while another urn contains 3 white and 5 black balls. Two balls are drawn from the first urn and put into the second and then a ball is drawn from the second urn. Find the probability that the ball drawn from the second urn is a white ball.

Ans- First urn: 10W, 3B
Second urn: 3W, 5B

Possible transfers:
2 white transferred
Probability = 10C2 / 13C2 = 45/78

Second urn becomes 5W,5B
P(white) = 5/10

1 white and 1 black transferred
Probability = (10×3)/78 = 30/78

Second urn becomes 4W,6B
P(white) = 4/10

2 black transferred
Probability = 3C2 / 13C2 = 3/78

Second urn becomes 3W,7B
P(white) = 3/10

Required probability
= (45/78×5/10) + (30/78×4/10) + (3/78×3/10)
= 225/780 + 120/780 + 9/780
= 354/780
= 59/130

–: End of Self Assessment and Revision on Probability Class 12 OP Malhotra ISC Maths Solutions Ch-18 :–

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