Question

A bag contains $6$ white and $4$ red balls. Three balls are drawn at random. What is the probability that one ball is red and the other two are white?

Answer 1

Given a bag contains $6$ white and $4$ red balls.

$\therefore$ The sample space is $S=6+4=10$

One ball is red and two balls are white.

Case$I:$ If red ball is drawn at first position and then the white balls

$P=\frac{6}{10}\times \frac{4}{9}\times \frac{3}{8}=\frac{6\times 4\times 3}{10\times 9\times 8}=\frac{1}{10}$

Case$II:$ If red ball is drawn at second position and then the white balls correspondingly.

$P=\frac{4}{10}\times \frac{6}{9}\times \frac{3}{8}=\frac{4\times 6\times 3}{10\times 9\times 8}=\frac{1}{10}$

Case$III:$ If red ball is drawn at third position

$P=\frac{4}{10}\times \frac{3}{9}\times \frac{6}{8}=\frac{4\times 3\times 6}{10\times 9\times 8}=\frac{1}{10}$

$\therefore$ Probability of one ball is red and two balls are white$=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}=\frac{3}{10}$