Question

Assertion (A) :There are two bags (i) & (ii), bag (i) contains $3$ red & $4$ black balls & bag (ii) contains $5$ red & $6$ black balls. If one ball is drawn at random from one of the bags and found to be red then the probability that it was drawn from bag (ii) is $35/68$.
Reason (R) : If $E_1$, & $E_2$ events constitute a partition of sample space $S$ & $A$ is any event of non zero probability, then $P(E_i/A)=\frac{P(E_i)P(A/E_i)}{P(A)},(i=1,2)$.

• A. Both (A) & (R) are individually true & (R) is correct explanation of (A).
• B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).
• C. (A) is true but (R) is false.
• D. (A) is false but (R) is true.

Answer 1

Answer: (A)

Let $E_1=$ event of choosing bag (i) & $E_2=$ event of choosing the bag (ii) & ${}^{\prime }{A}^{\prime }$ be the event of drawing red ball
$\therefore P(E_1)=P(E_2)=\frac{1}{2}$
$\therefore P(A/E_1)=\frac{3}{7},P(A/E_2)=\frac{5}{11}$
$\therefore P(E_2/A)=\frac{P(E_2)P(A/E_2)}{P(A)}$
$=\frac{\frac{1}{2}\times \frac{5}{11}}{\frac{1}{2}\times \frac{3}{7}+\frac{1}{2}\times \frac{5}{11}}=\frac{35}{68}$.