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1. 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. $\frac{2}{77}$

Solution:

Let S be the sample space
Then, n(S) = number of ways of drawing 3 balls out of 15

\begin{aligned} =^{15} C_{3} \\ = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} \\ = 455 \end{aligned}

Let E = event of getting all the 3 red balls

\begin{aligned} ∴ n (E) =^{5} C_{3} =^{5} C_{2} \\ = \frac{5 \times 4}{2 \times 1} = 10 \\ ∴ P (E) = \frac{n (E)}{n (S)} \\ = \frac{10}{455} \\ = \frac{2}{91} \end{aligned}

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