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1. A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

A. $\frac{10}{21}$

B. $\frac{11}{21}$

C. $\frac{2}{7}$

D. $\frac{5}{7}$

Solution:

Total number of balls
= (2 + 3 + 2)
= 7
Let S be the sample space
Then, n(S) = Number of ways of drawing 2 balls out of 7

\begin{aligned} n (S) =^{7} C_{2} \\ n (S) = \frac{(7 \times 6)}{(2 \times 1)} \\ n (S) = 21 \end{aligned}

Let E = Event of 2 balls, none of which is blue
∴ n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls

\begin{aligned} n (E) =^{5} C_{2} \\ n (E) = \frac{(5 \times 4)}{(2 \times 1)} \\ n (E) = 10 \\ ∴ P (E) = \frac{n (E)}{n (S)} = \frac{10}{21} \end{aligned}

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