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Q.
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.
Probability - Part 2
Solution:
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
$= \,{}^{3}C_{1}pq^{2} = \,{}^{3}C_{1}\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^{2} = \frac{3}{8}$.