In a bag there are 4 white, 4 red and 2 green balls. Two balls are drawn at random. What is the probability that at least one ball is of green colour ?
There are 4 white, 4 red and 2 green balls and two balls are drawn at random.
Total possible outcomes = Selection of 2 balls out of 10 balls
= $$C^{10}_2 = \frac{10 * 9}{1 * 2} = 45$$
Favourable outcomes = 1 green ball and 1 ball of other colour + 2 green balls
= $$C^2_1 \times C^8_1 + C^2_2$$
= 2*8 + 2 = 18
$$\therefore$$ Required probability = $$\frac{18}{45} = \frac{2}{5}$$
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