Question

A bag contains $4$ white, $3$ black and $2$ red balls. Balls are drawn from the bag one by one without replacement.
The probability that $5^{\text{th}}$ ball is black given that $4^{\text{th}}$ ball is red is

• A. $\frac{1}{8}$
• B. $\frac{3}{8}$
• C. $\frac{2}{11}$
• D. $\frac{3}{11}$

Answer 1

The correct option is B $\frac{3}{8}$

A bag contains $4$ white, $3$ black and $2$ red balls.
$A\to$ getting black ball on $5^{\text{th}}$ draw
$B\to$ getting red ball on $4^{\text{th}}$ draw
$P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P\left(B\right)},$ where $P\left(B\right)=\frac{2}{9}$
We need to find the probability when $4^{\text{th}}$ ball is red and $5^{\text{th}}$ ball is black i.e., $P(A\cap B)$
Number of ways of choosing $4^{\text{th}}$ ball $={}^2{C}_1$
Number of ways of choosing $5^{\text{th}}$ ball $={}^3{C}_1$
$P(A\cap B)=\frac{{}^2{C}_1\cdot {}^3{C}_1\cdot 7\cdot 6\cdot 5}{9\cdot 8\cdot 7\cdot 6\cdot 5}\Rightarrow P(A\cap B)=\frac{1}{12}\therefore P\left(\frac{A}{B}\right)=\frac{1/12}{2/9}=\frac{3}{8}$