Question

A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement, the probability of getting exactly one red ball is

• A. $\frac{45}{196}$
• B. $\frac{135}{392}$
• C. $\frac{15}{56}$
• D. $\frac{15}{29}$

Answer 1

Answer: (C)

Given that, red balls $=5$ and blue balls $=3$
Total number of balls $=5+3=8$
$P(\text{ red })=\frac{5}{8}$
$P(\text{ blue })=\frac{3}{8}$
Required probability is given by
$P(\text{ red })=P(\text{ red })P(\text{ blue })P(\text{ blue })+P(\text{ blue })P(\text{ red })P(\text{ blue })+P(\text{ blue })P(\text{ blue })P(\text{ red })$
$=\frac{5}{8}\times \frac{3}{7}\times \frac{2}{6}+\frac{3}{8}\times \frac{5}{7}\times \frac{2}{6}+\frac{3}{8}\times \frac{2}{7}\times \frac{5}{6}$
$=\frac{30}{336}+\frac{30}{336}+\frac{30}{336}$
$=\frac{90}{336}=\frac{15}{56}$