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 them 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.
Then, n(S) = 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 = event of getting all the 3 red balls.
$\therefore n\left(E\right){=}^5{C}_3{=}^5{C}_2=\frac{(5\times 4)}{(2\times 1)}=10$
$\therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{10}{455}=\frac{2}{91}$