Question

A bag contains $5$ white and $8$ red balls. Two successive drawings of $3$ balls are made such that the balls are replaced before the second drawing. Find the probability that the first drawing will give $3$ white and the second $3$ red balls.

• A. $\frac{210}{20449}$
• B. $\frac{150}{20449}$
• C. $\frac{140}{20449}$
• D. None of these

Answer 1

The correct option is C $\frac{140}{20449}$

First drawing will give $3$ white balls.

$\Rightarrow$ probability of drawing $3$ white balls(no replacement)

$p\left(w\right)$ = $\frac{5}{13}\times \frac{4}{12}\times \frac{3}{11}$

In second drawing $3$ white balls are placed and $3$ red balls are drawn.

$P\left(R\right)$ = $\frac{8}{13}\times \frac{7}{12}\times \frac{6}{11}$ Probability of drawing $3$ white and $3$ red balls $=P\left(W\right)\times P\left(R\right)$

$=\frac{5}{143}\times \frac{28}{143}=\frac{140}{20449}$