Question

A bag contains $6$ red, $4$ blue and $2$ yellow balls. Three balls are drawn one by one with replacement. Find the probability of getting exactly one red ball.

• A. $\frac{1}{4}$
• B. $\frac{3}{8}$
• C. $\frac{3}{4}$
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

$\frac{3}{8}$

Answer 2

Let $A$ be the event that ball drawn is red, then $p= \frac{6}{12} = \frac{1}{2}$, $q=1-\frac{1}{2}=\frac{1}{2}$ and $n = 3$ $\therefore$ Probability of drawing exactly $1$ red ball in $3$ trials $= \,^{3}C_{1}pq^{2} = \,^{3}C_{1}\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^{2} = \frac{3}{8}$.