Question

A box contains $12$ white and $12$ black balls. The balls are drawn at random from the box, one at a time with replacement. The probability that a white ball is drawn for the fourth time on the seventh draw, is

• A. $\frac{5}{64}$
• B. $\frac{27}{32}$
• C. $\frac{5}{32}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $\frac{5}{32}$

Due to replacement, each draw is identical and independent.
Getting the fourth white ball in the seventh draw means there should be $3$ white balls in the six draws.
This follows binomial distribution where $n=6$ and $p=P\left(\text{getting a white ball}\right)=\frac{12}{24}=\frac{1}{2}$

Required Probability $=P(\text{getting 3 white balls in first 6 draws and a white ball in}{7}^{\text{th}}\text{draw)}$
$={}^6{C}_3\cdot {\left(\frac{1}{2}\right)}^6.\frac{1}{2}=\frac{5}{32}$