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# Bag I contains 3 red and 4 black balls and bag II contains 4 red and 5 black balls. One ball is transferred from bag I to bag II and then a ball is drawn from bag II at random. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black. Comment
A)
Toolbox:
• According to Bayes Theorem, if $E_1, E_2, E_3.....E_n$ are a set of mutually exclusive and exhaustive events, then $P\left(\large \frac{E_i}{E}\right ) = \Large \frac{P\left(\frac{E}{E_i}\right ). P(E_i)} {\sum_{i=1}^{n} (P\left(\frac{E}{E_i}\right ).P(E_i))}$
Step 1:
Given Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball drawn is red in color.
Let $E_1$ be the event that a red ball is transferred from Bag I to Bag II, and $E_2$ be the event that a black ball is transferred from Bag I to Bag II. Let A be the event that the ball drawn is red.
$P (E_1) = \large\frac{3}{3+4} = \frac{3}{7}$
$P(E_2) = \large\frac{4}{4+3} = \frac{4}{7}$
Step 2:
$P \large(\frac{A}{E_1}) = \frac{5}{10} = \frac{1}{2}$
$P \large(\frac{A}{E_2}) = \frac{4}{10} = \frac{2}{5}$
$P\left(\large \frac{E_2}{E}\right ) = \Large \frac{P\left(\frac{E}{E_2}\right ). P(E_2)} {\sum_{i=1}^{4} (P\left(\frac{E}{E_i}\right ).P(E_i))}$.
$\Rightarrow \large\;\frac{\frac{4}{7}\;\times\;\frac{2}{5}}{\large\frac{3}{7}\;\times\;\frac{1}{2}\;+\;\frac{4}{7}\;\times\;\frac{2}{5}}=\large\;\frac{16}{31}$