Box I contains 30 cards numbered 1 to 30 and | vedclass

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(A) Let $B_{1}$ be the event that Box-$I$ is selected and $B_{2}$ be the event that Box-$II$ is selected.$P(B_{1}) = P(B_{2}) = \frac{1}{2}$.Let $E$ b

Vedclass · Accepted Answer

$\frac{8}{17}$

Vedclass · Answer

(A) Let $B_{1}$ be the event that Box-$I$ is selected and $B_{2}$ be the event that Box-$II$ is selected.$P(B_{1}) = P(B_{2}) = \frac{1}{2}$.Let $E$ be the event that the selected card is a non-prime number.In Box-$I$ (cards $1$ to $30$),the prime numbers are $\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29\}$ ($10$ primes). Thus,there are $30 - 10 = 20$ non-prime numbers. So,$P(E|B_{1}) = \frac{20}{30} = \frac{2}{3}$.In Box-$II$ (cards $31$ to $50$),the prime numbers are $\{31, 37, 41, 43, 47\}$ ($5$ primes). Thus,there are $20 - 5 = 15$ non-prime numbers. So,$P(E|B_{2}) = \frac{15}{20} = \frac{3}{4}$.Using Bayes' Theorem,the probability that the card was drawn from Box-$I$ given it is non-prime is:$P(B_{1}|E) = \frac{P(B_{1})P(E|B_{1})}{P(B_{1})P(E|B_{1}) + P(B_{2})P(E|B_{2})}$$P(B_{1}|E) = \frac{\frac{1}{2} 	imes \frac{2}{3}}{\frac{1}{2} 	imes \frac{2}{3} + \frac{1}{2} 	imes \frac{3}{4}} = \frac{\frac{1}{3}}{\frac{1}{3} + \frac{3}{8}} = \frac{\frac{1}{3}}{\frac{8+9}{24}} = \frac{1}{3} 	imes \frac{24}{17} = \frac{8}{17}$.

Box	Red	White	Black
$A$	$1$	$6$	$3$
$B$	$6$	$2$	$2$
$C$	$8$	$1$	$1$
$D$	$0$	$6$	$4$

Box $I$ contains $30$ cards numbered $1$ to $30$ and Box $II$ contains $20$ cards numbered $31$ to $50$. $A$ box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was drawn from Box $I$ is

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