Question

A bag contains 4 white, 5 red, and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of then are red, is

• A. $\frac{1}{22}$
• B. $\frac{3}{22}$
• C. $\frac{2}{91}$
• D. $\frac{2}{77}$

Answer 1

The correct option is C $\frac{2}{91}$

Let $S$ be the sample space

$\Rightarrow n\left(S\right)=$number of ways of drawing $3$ balls out of $15$

${15}_{C_3}=\frac{15\times 14\times 13}{3\times 2\times 1}=455$

Let $E$ be the event of getting all of the $3$ red balls.

$\therefore n\left(E\right)=5_{C_3}=5_{C_2}=\frac{5\times 4}{2\times 1}=10$

$\therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{10}{455}=\frac{2}{91}$