For two events A and B, if P(A)=P≤ft(A/B)=1/4 and P≤ft(B/A)=1/2, then

Correct Answer is : All of these

Correct Answer is : All of these

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CGPET 2019 Solved Paper

Section: Mathematics

Question : 120 of 150

Marks: +1, -0

For two events

A

B

P(A)=P\left(\frac{A}{B}\right)=\frac{1}{4}

P\left(\frac{B}{A}\right)=\frac{1}{2}

, then

Solution:

Since,

P(A)=P\left(\frac{A}{B}\right)=\frac{1}{4}

and

P\left(\frac{B}{A}\right)=\frac{1}{2}

\Rightarrow \frac{P(A \cap B)}{P(B)}=\frac{1}{4}

and

\frac{P(A \cap B)}{P(A)}=\frac{1}{2}

P(A \cap B)=\frac{1}{2} P(A)=\frac{1}{2} \times \frac{1}{4}=\frac{1}{8}

and

P(B)=4 P(A \cap B)=4 \times \frac{1}{8}=\frac{1}{2}

\because P(A \cap B)=\frac{1}{8}=P(A) P(B)

\therefore

Events

A

and

B

are independent.

\because P\left(\frac{A'}{B}\right)=1-P\left(\frac{A}{b}\right)=1-\frac{1}{4}=\frac{3}{4}

and

P\left(\frac{B'}{A'}\right)=\frac{P(A' \cap B')}{P(A')}=\frac{P((A \cup B)')}{P(A^{1})}

=\frac{1-P(A \cup B)}{1-P(A)}

=\frac{1-\{P(A)+P(B)-P(A \cap B)\}}{1-P(A)}

=\frac{1-\left\{\frac{1}{4}+\frac{1}{2}-\frac{1}{8}\right\}}{1-\frac{1}{4}}=\frac{1-\frac{5}{8}}{\frac{3}{4}}

=\frac{\frac{3}{8}}{\frac{3}{4}}=\frac{1}{2}

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