Question

There are $12$ white and $12$ red balls in a bag. Balls are drawn one by one with replacement from the bag. The probability that $7^{\mathrm{th}}$ drawn ball is $4^{\mathrm{th}}$ white, is

• A. $\frac{1}{4}$
• B. $\frac{5}{32}$
• C. $\frac{3}{16}$
• D. $\frac{5}{16}$

Answer 1

The correct option is B

$\frac{5}{32}$

Find the probability:

Let $\mathrm{p}$ be the probability of drawing a white ball and $\mathrm{q}$ be the probability of drawing a red ball.

$\begin{array}{rcl}\mathrm{p}& =& \frac{12}{24}\\ & \Rightarrow & \mathrm{p}=\frac{1}{2}\end{array}$

$\begin{array}{rcl}\mathrm{q}& =& \frac{12}{24}\\ & \Rightarrow & \mathrm{q}=\frac{1}{2}\end{array}$

We want to calculate probability that $7^{\mathrm{th}}$ drawn ball is $4^{\mathrm{th}}$ white, i.e. till the $6^{\mathrm{th}}$ attempt we should have $3$ white balls and white ball in $7^{\mathrm{th}}$ draw

Using the binomial distribution,

$\begin{array}{rcl}\mathrm{W}& =& \left[{}^6\mathrm{C}_3{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^{6-3}\right]\frac{1}{2}\left[{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}=\frac{\mathrm{n}!}{\mathrm{r}!\times (\mathrm{n}-\mathrm{r})!}\right]\\ & \Rightarrow & \mathrm{W}=\frac{6!}{3!\times 3!\times 2^3\times 2^3\times 2}\\ & \Rightarrow & \mathrm{W}=\frac{5}{32}\end{array}$

Hence option (B) is the correct option.