Mark the correct alternative in the following question:

A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement, then the probability that exactly two of the three balls were red, the first ball being red, is

$(a) \frac{1}{3} (b) \frac{4}{7} (c) \frac{15}{28} (d) \frac{5}{28}$

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Solution

$We have, The number of red balls = 5 and The number of blue balls = 3 Let R be the event of getting a red ball and B be the event of getting a blue ball . Now, P (Getting exactly two red balls of the three balls, the first ball being red) = P (R B | R) + P (B R | R) = \frac{4}{7} \times \frac{3}{6} + \frac{3}{7} \times \frac{4}{6} = \frac{2}{7} + \frac{2}{7} = \frac{4}{7}$

Hence, the correct alternative is option (b).

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Mark the correct alternative in the following question: A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement, then the probability that exactly two of the three balls were red, the first ball being red, is a 13 b 47 c 1528 d 528