Probability Question 48
Question: Let A and B are two independent events. The probability that both A and B occur together is 1/6 and the probability that neither of them occurs is 1/3. The probability of occurrence of A is
[RPET 2000]
Options:
A) 0 or 1
B) $ \frac{1}{2} $ or $ \frac{1}{3} $
C) $ \frac{1}{2} $ or $ \frac{1}{4} $
D) $ \frac{1}{3} $ or $ \frac{1}{4} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ P(A\cap B)=\frac{1}{6} $ and $ P(A^{c}\cap B^{c})=\frac{1}{3} $
Now $ P{{(A\cup B)}^{c}}=P(A^{c}\cap B^{c})=\frac{1}{3} $
Therefore $ 1-P(A\cup B)=\frac{1}{3} $ $ \Rightarrow P(A\cup B)=\frac{2}{3} $
But $ P(A\cup B)=P(A)+P(B)-P(A\cap B) $ $ \Rightarrow P(A)+P(B)=\frac{5}{6} $ -..(i)
$ \because $ A and B are independent events $ P(A\cap B)=P(A)P(B) $
Therefore $ P(A)P(B)=\frac{1}{6} $ $ {{[P(A)-P(B)]}^{2}}={{[P(A)+P(B)]}^{2}}-4P(A)P(B) $
$ =\frac{25}{36}-\frac{4}{6}=\frac{1}{36} $
$ \Rightarrow $ $ P(A)-P(B)=\pm \frac{1}{6} $ ……(ii)
Solving (i) and (ii), we get $ P(A)=\frac{1}{2} $ or $ \frac{1}{3}. $
BETA
