Question

A bag contains $2$ white and $4$ black balls. A ball is drawn $5$ times with replacement. The probability that at least $4$ of the balls drawn are white is

• A. $\frac{8}{141}$
• B. $\frac{10}{243}$
• C. $\frac{11}{243}$
• D. $\frac{8}{41}$

Answer 1

The correct option is C

$\frac{11}{243}$

Explanation for correct answer:

Step-1: Finding probability using classical approach.

Given, A bag contains $2$ white and $4$ black balls.

Total number of ball is $2+6=6$.

The probability that of getting white ball is $\frac{2}{6}=\frac{1}{3}$

The probability that of getting white ball is $\frac{4}{6}=\frac{2}{3}$

Step-2: Finding probability that at least $4$ of the balls drawn are white by using binomial distribution.

A ball drawn $5$ times $n=5$.

The probability that at least $4$ of the balls drawn are white is

Probability at $x=4$$+$Probability at $x=5$

$={}^{5}C_4\times {\left(\frac{1}{3}\right)}^4\times {\left(\frac{2}{3}\right)}^1+{}^{5}C_5\times {\left(\frac{1}{3}\right)}^5\times {\left(\frac{2}{3}\right)}^0$

Step-3: Apply formula $\left({}^{n}C_r=\frac{n!}{\left(n-r\right)!\times r!}\right)$.

$$\begin{gathered} =\frac{5!}{4!\times \left(5-4\right)!}\times \frac{1}{81}\times \frac{2}{3}+1\times \frac{1}{243}\times 1 \\ =\frac{5\times 4!}{4!\times 1!}\times \frac{2}{243}+\frac{1}{243} \\ =\frac{10+1}{243} \\ =\frac{11}{243} \end{gathered}$$

Hence, the correct answer is option $\left(C\right)$.