Question
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?
Answer: Option A
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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
= 7C2
= \(\frac{(7\times6)}{(2\times1)}\)
=21.
Let E = Event of drawing 2 balls, none of which is blue.
So, n(E) = = Number of ways of drawing 2 balls out of (2 + 3) balls.
= 5C2
= \(\frac{(5\times4)}{(2\times1)}\)
=10.
So, \(P(E) = \frac{n(E)}{n(S)}=\frac{10}{21}\)
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