Practicing ISC Class 12 OP Malhotra Solutions Chapter 18 Probability Ex 18(a) is the ultimate need for students who intend to score good marks in examinations.
S Chand Class 12 ICSE Maths Solutions Chapter 18 Probability Ex 18(a)
Question 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 ?
Solution:
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)
= \(\frac { 8 }{ 20 }\) + \(\frac { 5 }{ 20 }\) = \(\frac { 13 }{ 20 }\)
Question 2.
If P (A) = \(\frac { 1 }{ 5 }\), P(B) = \(\frac { 2 }{ 3 }\) and P (A ∩ B) = \(\frac { 4 }{ 15 }\), are A and B exhaustive events.
Solution:
Given P (A) = \(\frac { 1 }{ 5 }\), P (B) = \(\frac { 2 }{ 3 }\)
P (A ∩ B) = \(\frac { 4 }{ 15 }\)
Here P(A∪B) = P (A) + P (B) – P (A ∩ B)
= \(\frac{1}{5}+\frac{2}{3}-\frac{4}{15}=\frac{3+10-4}{15}=\frac{9}{15}=\frac{3}{5} \neq 1\)
∴ A and B are not exhaustive events.
Question 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.
Solution:
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) = \(\frac { 1 }{ 6 }\) and P(B) = \(\frac { 3 }{ 6 }\)
(i) Thus P (Rolling a 5 or odd number comes) 13 4 2
= P(A) + P(B) = \(\frac{1}{6}+\frac{3}{6}=\frac{4}{6}=\frac{2}{3}\)
(ii) When 2 dice are rolled
Then total no. of outcomes = 62 = 36
E : event of getting atleast one 4 when 2 dice rolling.
∴ favourable outcomes are {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (1, 4), (2, 4), (3, 4), (5, 4), (6, 4)}
Thus total no. of favourable outcomes = 11
∴ required probability = \(\frac { 11 }{ 36 }\)
Question 4.
The probability that Dimple goes to the local shop is \(\frac { 3 }{ 7 }\). The probability that she does not cycle is \(\frac { 8 }{ 21 }\). The probability that she goes to the shop and cycles is \(\frac { 16 }{ 35 }\).
(i) What is the probability that she cycles?
(ii) What is the probability that she cycles or goes to the shop?
Solution:
Given P (Dimple goes a local shop) = \(\frac { 3 }{ 7 }\)
and P (does not goes by cycles) = \(\frac { 8 }{ 21 }\)
∴ P (she goes by cycle) = 1 – \(\frac{8}{21}=\frac{13}{21}\)
Thus P (she goes to shop and by cycles) = \(\frac { 16 }{ 35 }\)
(i) required probability that she cycles = \(\frac { 13 }{ 21 }\)
(ii) required probability = \(\frac{3}{7}+\frac{13}{21}-\frac{16}{35}=\frac{62}{105}\)
Question 5.
A fair die is thrown. What is the probability that either an odd number or a number greater than 4 will turn up?
Solution:
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)
= \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}\)
= \(\frac{3}{6}+\frac{2}{6}-\frac{1}{6}=\frac{4}{6}=\frac{2}{3}\)
Question 6.
It is given that for two events A and B, P(A) = \(\frac { 3 }{ 8 }\), P(A∪B)= \(\frac { 11 }{ 16 }\) and P(A∩ B) = \(\frac { 3 }{ 16 }\). Find P(B).
Solution:
Given P (A) = \(\frac { 3 }{ 8 }\) ; P (A ∪ B) = \(\frac { 11 }{ 16 }\)
and P (A ∩ B) = \(\frac { 3 }{ 16 }\)
We know that P(A∪B) = P (A) + P (B) – P (A ∩ B)
⇒ \(\frac{11}{16}=\frac{3}{8}+P(B)-\frac{3}{16}\)
⇒ P(B) = \(\frac{11}{16}+\frac{3}{16}-\frac{3}{8}\)
⇒ P(B) = \(\frac{11+3-6}{16}=\frac{8}{16}=\frac{1}{2}\)
Question 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
Solution:
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
Question 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
Solution:
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)
∴ \(P(\bar{A})+P(\bar{B})\) = 1 – P (A) + 1 – P (B)
= 2 – [P (A) + P (B)]
= 2 – 1.1 = 0.9
Question 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
Solution:
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
Question 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’)
(d) None of these
Solution:
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’)
Question 11.
If A and B are events of a random experiment such that P(A ∪ B) = \(\frac { 4 }{ 5 }\), P(A’ ∪B’) = \(\frac { 7 }{ 10 }\) and P(B) = \(\frac { 2 }{ 5 }\), then P(A) is equal to
(a) \(\frac { 3 }{ 5 }\)
(b) \(\frac { 7 }{ 10 }\)
(c) \(\frac { 8 }{ 10 }\)
(d) \(\frac { 9 }{ 10 }\)
Solution:
Given P(A ∪ B) = \(\frac { 4 }{ 5 }\) ; P(A’ ∪B’) = \(\frac { 7 }{ 10 }\) ; P(B) = \(\frac { 2 }{ 5 }\)
Now P (A’ ∪ B’) = \(\frac { 7 }{ 10 }\)
⇒ \(\frac { 7 }{ 10 }\) = P[(A ∩ B)’]
⇒ \(\frac{7}{10}=1-P(A \cap B)=1-[P(A)+P(B)-P(A \cup B)]\)
⇒ \(\frac{7}{10}=1-P(A)-\frac{2}{5}+\frac{4}{5}\)
⇒ \(\frac{7}{10}=\frac{7}{5}-P(A)\)
⇒ \(P(A)=\frac{7}{5}-\frac{7}{10}=\frac{7}{10}\)
Question 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
Solution:
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]