Question

A bag contains $8$ red and $5$ white balls. Two successive draws of $3$ balls are made without replacement. The probability that the first draw will produce $3$ white balls and the second $3$ red ball is

• A. $\frac{5}{223}$
• B. $\frac{6}{257}$
• C. $\frac{7}{429}$
• D. $\frac{3}{548}$

Answer 1

Answer: (C)

$\frac{7}{429}$

A bag contain $8$ red and $5$ white balls

$\Rightarrow P(A\cap B)=P\left(A\right)P\left(\frac{B}{A}\right)$

A is the event such that the first drawing will give $3$ white balls.

B is the event such that the second drawing will give $3$ blue balls.

$\Rightarrow P\left(A\right)=\frac{\frac{5!}{3!2!}}{\frac{13!}{3!10!}}=\frac{5!10!}{2!3!}=\frac{5}{143}$

$\Rightarrow P\left(\frac{B}{A}\right)=\frac{\frac{8!}{3!5!}}{\frac{10!}{3!7!}}=\frac{8!7!}{5!10!}=\frac{7}{15}$

$\Rightarrow P(A\cap B)=P\left(A\right)\left(\frac{B}{A}\right)=\frac{5}{143}\times \frac{7}{15}=\frac{7}{429}=0.01632$

Hence, the answer is $\frac{7}{429}.$