Question

Bag $I$ contains $3$ red and $4$ black balls and Bag $II$ contains $4$ red and $5$ black. One ball is transferred from bag $I$ to bag $II$ and then a ball is drawn from bag $II$ . The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Answer 1

Hint : First find the number of balls left in each bag after the transfer is done. Then use Bayes’ Theorem on the conditional probability to solve this question.

Complete step-by-step answer :
Let’s consider the events ${E_1}$ and ${E_2}$ as follows.
${E_1} =$ ball transferred from a bag $I$ to bag $II$ is red.
${E_2} =$ Ball transferred from bag $I$ to bag $II$ is black.
$A =$ Ball drawn from bag $II$ is red
$P({E_1})$ is the probability of transferring red ball from bag $I$ to bag $II$
$\Rightarrow P({E_1}) = \dfrac{3}{7}$ (since, bag $I$ contains $3$ red balls and $4$ black balls)
$P({E_2})$ is the probability of transferring black ball from bag $I$ to bag $II$
$\Rightarrow P({E_2}) = \dfrac{4}{7}$ (since, Bag $I$ contains $3$ red and $4$ black balls)
Red balls are added to bag $II$ .
Therefore, the total number of balls in bag $II$ is $10$
And number of red balls is $4 + 1 = 5$
By using the formula of conditional probability, we can write
Probability of drawing red ball from bag $II$ when red ball is already transferred from bag $I$ to bag $II$ is given by $P(A|{E_1})$
$\Rightarrow P(A|{E_1}) = \dfrac{5}{{10}}$
$\Rightarrow P(A|{E_1}) = \dfrac{1}{2}$
Now, let is assume that a black ball is added to bag $II$
Therefore, the total number of balls in bag $II$ is 10
And the number of red balls is 4
Then, probability of drawing red ball from bag $II$ when black ball is already transferred from bag $I$ to bag $II$ is given by $P(A|{E_2})$
$\Rightarrow P(A|{E_1}) = \dfrac{4}{{10}}$
$\Rightarrow P(A|{E_1}) = \dfrac{2}{5}$
Probability of transferring a black ball when the ball drawn from bag $II$ is found to be red is given by $P({E_2}|A)$
Using Bayes’ theorem, we can write
$P({E_2}|A) = \dfrac{{P({E_2})P(A|{E_2})}}{{P({E_1})P(A|{E_1}) + P({E_2})P(A|{E_2})}}$
$= \dfrac{{\dfrac{4}{7} \times \dfrac{2}{5}}}{{\dfrac{3}{7} \times \dfrac{1}{2} + \dfrac{4}{7} \times \dfrac{2}{5}}}$
Simplifying it, we get
$P({E_2}|A) = \dfrac{{\dfrac{8}{{35}}}}{{\dfrac{1}{7}\left( {\dfrac{3}{2} + \dfrac{8}{5}} \right)}}$
$= \dfrac{{\dfrac{8}{{35}} \times 7}}{{\dfrac{{15 + 16}}{{10}}}}$
$= \dfrac{{\dfrac{8}{5}}}{{\dfrac{{31}}{{10}}}}$
$= \dfrac{8}{5} \times \dfrac{{10}}{{31}}$ $\left( {\because \dfrac{{\dfrac{a}{b}}}{{\dfrac{p}{q}}} = \dfrac{a}{b} \times \dfrac{q}{p}} \right)$
$= \dfrac{{8 \times 2}}{{31}}$
$\Rightarrow P({E_2}|A) = \dfrac{{16}}{{31}}$
Therefore, probability of transferring a black ball from bag $I$ to bag $II$ when the ball drawn from bag $II$ is found to be red is given by $P({E_2}|A)$

Note : Conditional probability is tricky to understand. $P(A|B)$ is the probability of occurrence of event A when event B has already happened or true. $P(B|A)$ is the probability of occurrence of event B when event A has already happened or true.
$P(A|B) \ne P(B|A)$ So, be careful while using conditional probability.