Question

A box contains $24$ identical balls, of which $12$ are white and $12$ are black. 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 $4th$ time on the $7th$ 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

$\frac{5}{32}$

An explanation for the correct option:

Step 1: Find the probability for a white ball is drawn and find the probability for a black ball is drawn

We have been given that, a box contains $24$ identical balls, of which $12$ are white and $12$ are black. The balls are drawn at random from the box one at a time with replacement.

We need to find the probability that a white ball is drawn for the $4th$ time on the $7th$ draw.

Since there is a box contains $24$ identical balls, of which $12$ are white and $12$ are black.

Let, $p:$The probability that white ball is drawn

$$\begin{gathered} \Rightarrow p=\frac{12}{24} \\ \Rightarrow p=\frac{1}{2} \end{gathered}$$

$q:$The probability that black ball is drawn

$$\begin{gathered} \Rightarrow q=\frac{12}{24} \\ \Rightarrow q=\frac{1}{2} \end{gathered}$$

Step 2: Find the required probability.

The probability that a white ball is drawn for the $4th$ time on the $7th$ draw.

This means the white ball is drawn $3$ times in the first $6$ draws.

So, the required probability would be,

$$\begin{gathered} \Rightarrow P={}^{6}C_3\times {\left(\frac{1}{2}\right)}^3\times {\left(\frac{1}{2}\right)}^3\times \left(\frac{1}{2}\right) \\ \Rightarrow P=\frac{6!}{3!(6-3)!}\times {\left(\frac{1}{2}\right)}^7 \\ \Rightarrow P=20\times \frac{1}{128} \\ \Rightarrow P=\frac{5}{32} \end{gathered}$$

Therefore, option (C) is the correct answer.