Question

Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. 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 black in colour. Then the probability, that the transferred ball is red, is :

• A. $\frac{4}{9}$
• B. $\frac{5}{18}$
• C. $\frac{1}{6}$
• D. $\frac{3}{10}$

Answer 1

Answer: (B)

$\frac{5}{18}$

Answer 2

Let E→Ball drawn from Bag II is black.
ER→Bag I to Bag II red ball transferred.
EB→Bag I to Bag II black ball transferred.
EW→Bag I to Bag II white ball transferred.

$P(\frac{E_R}{E})=\frac{P(\frac{E}{ER})⋅P(ER) }{ P(\frac{E}{ER})P(ER)+P(\frac{E}{EB})⋅P(EB)+P(\frac{E}{EW})P(EW)}$

Here,
P(ER)=$\frac{3}{10}$, P(EB)=$\frac{4}{10}$, P(EW)=$\frac{3}{10}$
and
$P(\frac{E}{E_R})=\frac{5}{10},P(\frac{E}{E_B})=\frac{5}{10},P(\frac{E}{E_W})=\frac{5}{10}$

$∴P(\frac{E_R}{E})=\frac{\frac{15}{100} }{\frac{ 15}{100}+\frac{24}{100}+\frac{15}{100}}$
=$\frac{15}{54}$
$=\frac{5}{18}$
So, the correct option is (B): $\frac{5}{18}$