Question

A bag contains $5$ white and $8$ red balls. Two successive drawings of $3$ balls are made such that the balls are not replaced before the second draw. Find the probability that the first drawing will give $3$ white and the second $3$ red balls.

• A. $\frac{5}{429}$
• B. $\frac{9}{429}$
• C. $\frac{7}{429}$
• D. None of these

Answer 1

The correct option is C $\frac{7}{429}$

A bag contains $5$ white and $8$ red balls.

Total number of balls $=13$.

Two successive drawings of $3$ balls are made such that the balls are not replaced before the second draw.

We have to find the probability that the first drawing will give $3$ white and the second $3$ red balls.

The probability of first three balls are white in first draw, with no replacement is $\frac{5}{13}\times \frac{4}{12}\times \frac{3}{11}=\frac{60}{1716}$

Now the second draw, the probability that all are red is $\frac{8}{10}\times \frac{7}{9}\times \frac{6}{8}=\frac{336}{720}$

Thus the probability that the first drawing will give $3$ white and the second $3$ red balls is $\frac{60}{1716}\times \frac{336}{720}=\frac{7}{429}$.