A bag A contains 2 white and 3 red balls and | vedclass

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(D) Let $E_1$ be the event that the ball is drawn from bag $A$,$E_2$ be the event that it is drawn from bag $B$,and $E$ be the event that the drawn ba

Vedclass · Accepted Answer

$\frac{25}{52}$

Vedclass · Answer

(D) Let $E_1$ be the event that the ball is drawn from bag $A$,$E_2$ be the event that it is drawn from bag $B$,and $E$ be the event that the drawn ball is red.We need to find $P(E_2|E)$.Since both bags are equally likely to be selected,we have $P(E_1) = P(E_2) = \frac{1}{2}$.Also,the probability of drawing a red ball from bag $A$ is $P(E|E_1) = \frac{3}{5}$,and from bag $B$ is $P(E|E_2) = \frac{5}{9}$.By Bayes' theorem:$P(E_2|E) = \frac{P(E_2) \cdot P(E|E_2)}{P(E_1) \cdot P(E|E_1) + P(E_2) \cdot P(E|E_2)}$$P(E_2|E) = \frac{\frac{1}{2} \cdot \frac{5}{9}}{\frac{1}{2} \cdot \frac{3}{5} + \frac{1}{2} \cdot \frac{5}{9}}$$P(E_2|E) = \frac{\frac{5}{18}}{\frac{3}{10} + \frac{5}{18}} = \frac{\frac{5}{18}}{\frac{27 + 25}{90}} = \frac{5}{18} \cdot \frac{90}{52} = \frac{5 \cdot 5}{52} = \frac{25}{52}$.

$Face$	$1, 2, 3, 4, 5, 6$
$Probability$	$0.1, 0.24, 0.19, 0.18, 0.15, 0.14$

$A$ bag $A$ contains $2$ white and $3$ red balls and bag $B$ contains $4$ white and $5$ red balls. One ball is drawn at random from a randomly chosen bag and is found to be red. The probability that it was drawn from bag $B$ is:

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