Bag I contains 3 red and 4 black balls and bag II contains 4 red and 4 black balls. Two balls are transferred at random from bag I to bag II and then a ball is drawn from bag II. The ball so drawn is found to be red colour, then what is the probability that the transferred balls were both black.

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$\frac{4}{17}$

$\frac{5}{17}$

$\frac{16}{17}$

$\frac{15}{17}$

answer is A.

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Detailed Solution

Event $E_{1} = {\begin{cases} d r a w n t w o b a l l s f r o m \\ b a g I i s b o t h a r e b l a c k \end{cases}}$

$P (E_{1}) = \frac{{}^{4}C_{2}}{{}^{7}C_{2}} = \frac{2}{7}$

Event $E_{2} = {b o t h a r e r e d} \Rightarrow P (E_{2}) = \frac{{}^{3}C_{2}}{{}^{7}C_{2}} = \frac{1}{7}$

Event $E_{3} = {o n e b l a c k, o n e r e d} \Rightarrow P (E_{3}) = \frac{{}^{3}C_{1} \times {}^{4}C_{1}}{{}^{7}C_{2}} = \frac{4}{7}$

Let the event $= {After transforming balls from I to II the drawn ball from II is red}$

$P (A / E_{1}) = \frac{{}^{4}C_{1}}{{}^{10}C_{1}} = \frac{2}{5}, P (A / E_{2}) = \frac{{}^{6}C_{1}}{{}^{10}C_{1}} = \frac{3}{5}$

$P (A / E_{3}) = \frac{5 C_{1}}{10 C_{1}} = \frac{1}{2}$

$\begin{array}{l} ∴ P (E_{1} / A) = \frac{P (E_{1}) P (A / E_{1})}{P (E_{1}) P (A / E_{1}) + P (E_{2}) P (A / E_{2}) + P (E_{3}) P (A / E_{3})} \\ = \frac{\frac{2}{7} (\frac{2}{5})}{\frac{2}{7} (\frac{2}{5}) + \frac{1}{7} (\frac{3}{5}) + \frac{4}{7} (\frac{1}{2})} \\ = \frac{4}{4 + 3 + 10} = \frac{4}{17} \end{array}$