Concept of Probability Class 12 OP Malhotra Exe-18A ISC Maths Solutions Ch-18. In this article you would learn about the concepts of probability. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

Concept of Probability Class 12 OP Malhotra Exe-18A ISC Maths Solutions

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-18	Probability
Writer	OP Malhotra
Exe-18(a)	the concepts of probability

The concepts of probability

 Probability Class 12 OP Malhotra Exe-18A Solutions

Que-1: A bag contains 5 red balls, 8 black balls and 7 yellow balls. What is the probability of drawing either a red ball or a black ball from the bag ?

Sol: No. of black balls in a bag = 8
No. of red balls in a bag = 5
No. of yellow balls in a bag = 7
Total no. of balls in the bag = 5 + 8 + 7 = 20
∴ required probability = P (getting a black ball) + P (getting a red ball)
= 8/20 + 5/20 = 13/20

Que-2: If P (A) = 1/5, P(B) = 2/3 and P (A ∩ B) = 4/15, are A and B exhaustive events.

Sol: Given P (A) = 1/5, P (B) = 2/3
P (A ∩ B) = 4/15
Here P(A∪B) = P (A) + P (B) – P (A ∩ B)
= 1/5+2/3−4/15=3+10−4/15=9/15=3/5≠1
∴ A and B are not exhaustive events.

Que-3: Find each probability on a die
(i) rolling a 5 or an odd number.
(ii) rolling at least one 4 when rolling 2 dice.

Sol: When a die is rolled then S = {1, 2, 3, 4, 5, 6}
Total no. of outcomes = 6
A : event that 5 has comes on die
B : odd number comes
∴ P (A) = 1/6 and P(B) = 3/6
(i) Thus P (Rolling a 5 or odd number comes) 13 4 2
= P(A) + P(B) = 1/6+3/6=4/6=2/3

Que-4: The probability that Dimple goes to the local shop is 3/7. The probability that she does not cycle is 8/21. The probability that she goes to the shop and cycles is 16/35.
(i) What is the probability that she cycles?
(ii) What is the probability that she cycles or goes to the shop?

Sol: Given P (Dimple goes a local shop) = 3/7
and P (does not goes by cycles) = 8/21
∴ P (she goes by cycle) = 1 – 8/21=13/21
Thus P (she goes to shop and by cycles) = 16/35
(i) required probability that she cycles = 13/21
(ii) required probability = 3/7+13/21−16/35=62/105

Que-5: A fair die is thrown. What is the probability that either an odd number or a number greater than 4 will turn up?

Sol: When a fair dice is thrown.
Then S = {1,2, 3,4, 5,6}
A: event of getting an odd number ={1,3,5}
B : event that a number > 4 = {5, 6}
∴ n (A) = 3 ; n (B) = 2 i.e. n (S) = 6
∴ A ∩ B = {5} ⇒ n (A ∩B) = 1
(i) required probability = P(A ∪ B)
= P (A) + P (B) – P (A ∩ B)
= n( A)/n( S)=n( B)/n( S)=n( A∩B)/n( S)
= 3/6+2/6−1/6=4/6=2/3

Que-6: It is given that for two events A and B, P(A) = 3/8, P(A∪B)= 11/16 and P(A∩ B) = 3/16. Find P(B).

Sol: Given P (A) = 3/8 ; P (A ∪ B) = 11/16
and P (A ∩ B) = 3/16
We know that P(A∪B) = P (A) + P (B) – P (A ∩ B)
⇒ 11/16=3/8+P(B)−3/16
⇒ P(B) = 11/16+3/16−3/8
⇒ P(B) = 11+3−6/16=8/16=1/2

Que-7: If A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35, then P(A ∩ B’) = is equal to
(a) 0.12
(b) 0.19
(c) 0.34
(d) 0.88

Sol: Given P (A) = 0.54, P (B) = 0.69,
P (A ∩ B) = 0.35
P(A∩B’) = P (A) – P(A ∩B)
= 0.54 – 0.35 = 0.19

Que-8: If P(A ∪ B) = 0.8 and P(A ∩ B) = 0.3, then P(A) + P(B) is equal to
(a) 0.3
(b) 0.5
(c) 0.8
(d) 0.9

Sol: Given P(A∪B) = 0.8, P (A ∩ B) = 0.3
We know that P(A∪B) = P (A) + P (B) – P (A ∩ B)
⇒ P (A) + P (B) = P (A ∪ B) + P (A ∩ B)
= 0.8 + 0.3 = 1.1 …(1)
∴ P(A¯)+P(B¯) = 1 – P (A) + 1 – P (B)
= 2 – [P (A) + P (B)]
= 2 – 1.1 = 0.9

Que-9: If A and B are mutually exclusive events such that P(A) = 0.25, P(B) = 0.4, then P(A’ ∩ B’) is equal to
(a) 0.35
(b) 0.45
(c) 0.55
(d) 0.65

Sol: Since A and B are mutually exclusive events
∴ P(A∩B) = 0 …(1)
Given P (A) = 0.25 ; P (B) = 0.4
∴ P (A’ ∩ B’) = P [(A ∪ B)’]
= 1 – P (A ∪ B)
= 1 – [P(A) + P(B) – P(A ∩B)]
= 1 – [0.25 + 0.40 – 0] = 1 – 0.65 = 0.35

Que-10: If A and B are two mutually exclusive events, then
(a) P(A) < P(B’)
(b) P(A) < P(B)
(c) P(A) > P(B’)

Sol: Since A and B are mutually exclusive events
∴ P(A ∩ B) = 0 …(1)
Thus, P (A ∪B) = P (A) + P (B) ≤ 1 [using (1)]
⇒ P (A) < 1 – P (B) = P (B’)

Que-11: If A and B are events of a random experiment such that P(A ∪ B) = 4/5, P(A’ ∪B’) = 7/10 and P(B) = 2/5, then P(A) is equal to
(a) 35
(b) 7/10
(c) 8/10
(d) 9/10

Sol: Given P(A ∪ B) = 4/5 ; P(A’ ∪B’) = 7/10 ; P(B) = 2/5
Now P (A’ ∪ B’) = 7/10
⇒ 7/10 = P[(A ∩ B)’]
⇒ 7/10=1−P(A∩B)=1−[P(A)+P(B)−P(A∪B)]
⇒ 7/10=1−P(A)−2/5+4/5
⇒ 7/10=7/5−P(A)
⇒ P(A)=7/5−7/10=7/10

Que-12: The probability of event A occuring is 0.5 and of B occuring is 0.3. If A and B are mutually exclusive events, then the probability of neither A not B occuring is
(a) 0.5
(b) 0.6
(c) 0.7
(d) None of these

Sol: Given P (A) = 0.5 ; P (B) = 0.3
∴ required probability = P (neither A nor B) = P (A’ ∩ B’) = P [(A ∪ B)’]
= 1 – P (A ∪ B) = 1 – [P (A) + P (B) – P (A ∩ B)]
= 1 – 0.5 – 0.3 – 0 = 0.2
[since A and B are mutually exclusive events ∴ P (A ∩ B) = 0]

–: End of Concept of Probability Class 12 OP Malhotra Exe-18A ISC Maths Solutions Ch-18 Questions :–

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