Question

Coloured balls are distributed in three bags as shown in the following table :

$A$ bag is selected at random and then two balls are drawn from the selected bag. Find the probability that the balls drawn are black and blue.

• A. $\frac{163}{630}$
• B. $\frac{161}{630}$
• C. $\frac{169}{630}$
• D. $\frac{167}{630}$

Answer 1

Answer: (C)

Let $E_1$, $E_2$, $E_3$ and $A$ be the events defined as follows :
$E_1 =$ bag $I$ is selected
$E_2 =$ bag $II$ is selected
$E_3 =$ bag $III$ is selected
$A =$ one black and one blue ball has been drawn from the selected bag.
As the bags are selected at random,
$P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$
Two balls are drawn randomly from the selected bag.
$P(A|E_1) =$ probability of drawing one black and one blue ball when bag $I$ has been selected
$= \frac{^{2}C_{1} \times ^{3}C_{1}}{^{6}C_{2}} = \frac{2\times3}{15} = \frac{2}{5}$
$P(A|E_2) =$ probability of drawing one black and one blue ball when bag $II$ has been selected
$= \frac{^{4}C_{1} \times ^{1}C_{1}}{^{7}C_{2}} = \frac{4\times1}{21} = \frac{4}{21}$
$P(A|E_3) =$ probability of drawing one black and one blue ball when bag $III$ has been selected
$= \frac{^{3}C_{1} \times ^{2}C_{1}}{^{8}C_{2}} = \frac{3\times2}{28} = \frac{3}{14}$
By using law of total probability, we get
$P(A) = P(E_1) P(A|E_1) + P(E_2) P(A|E_2) + P(E_3) P(A|E_3)$
$= \frac{1}{3}\cdot\frac{2}{5}+\frac{1}{3}\cdot\frac{4}{21}+\frac{1}{3}\cdot\frac{3}{14}$
$= \frac{1}{3}\left[\frac{2}{5}+\frac{4}{21}+\frac{3}{14}\right]$
$= \frac{1}{3}\cdot\frac{169}{210} = \frac{169}{630}$