Question

A bag contain 5 white and 8 red balls. two drawings of 3 balls are made such that (a) the balls are replaced before the second trial and (b) the balls are not replaced before the second trial. Find the probability that the first drawing will give 3 white and the second, 3 red balls in each case

Answer 1

Total number of balls is 13.

First drawing:

Number all possible outcomes selecting 3 balls:

$C(13,3)=\frac{13!}{10!3!}=286$

Number of selecting 3 white ball is combination:

$C(5,3)=\frac{5!}{3!2!}=10$

P(3 white balls)$=\frac{10}{286}=\frac{5}{143}$

Second drawing:

Number of selecting 3 red balls is combination:

$C(8,3)=\frac{8!}{3!5!}=56$

a) with replacement

P(3 red balls)$=\frac{C(8,3)}{C(13,3)}=\frac{56}{286}=\frac{28}{143}$

So P(3 white and 3 red balls):

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

b) without replacement

After drawing 3 white balls bag contains 2 white and 8 red balls. 10 in total.

Hence number of selecting 3 red balls is combination:

$C(10,3)=\frac{10!}{3!7!}=120$

P(3 red balls)$=\frac{C(8,2)}{C(10,3)}=\frac{56}{120}=\frac{7}{15}$

so P(3 white and 3 red balls):

$\frac{5}{143}\cdot\frac{7}{15}=\frac{7}{429}$

Answer:

a) $\frac{140}{20449}$

b) $\frac{7}{429}$