Question 5

In a bag there are 15 red balls and 10 green balls. Three balls are selected at random. The probability of selecting 2 red balls and 1 green ball is :

Solution

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Number of ways of selecting 2 red balls and 1 green ball = $$15_{C_2}.\ 10_{C_1}$$

Number of ways of selecting 3 balls = $$25_{C_3}$$

Probability of selecting 2 red balls and 1 green ball = $$\ \frac{15_{C_2}.10_{C_1}}{25_{C_3}}$$ = $$\frac{21}{46}$$

Answer is option A.

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