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
$=\frac{{}^2{C}_1{\times }^3{C}_1}{{}^6{C}_2}=\frac{2\times 3}{15}=\frac{2}{5}$
$P(A|E_2)=$ probability of drawing one black and one blue ball when bag $II$ has been selected
$=\frac{{}^4{C}_1{\times }^1{C}_1}{{}^7{C}_2}=\frac{4\times 1}{21}=\frac{4}{21}$
$P(A|E_3)=$ probability of drawing one black and one blue ball when bag $III$ has been selected
$=\frac{{}^3{C}_1{\times }^2{C}_1}{{}^8{C}_2}=\frac{3\times 2}{28}=\frac{3}{14}$
By using law of total probability, we get
$P(A)=P(E_1)P(A|E_1)+P(E_2)P(A|E_2)+P(E_3)P(A|E_3)$
$=\frac{1}{3}\cdot \frac{2}{5}+\frac{1}{3}\cdot \frac{4}{21}+\frac{1}{3}\cdot \frac{3}{14}$
$=\frac{1}{3}\left[\frac{2}{5}+\frac{4}{21}+\frac{3}{14}\right]$
$=\frac{1}{3}\cdot \frac{169}{210}=\frac{169}{630}$