Statistics And Probability Question 570

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} $                     
    

Þ $ 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) $
Þ $ 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}. $