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Question
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
Options
\[ \frac{1}{3}\]
\[ \frac{4}{7}\]
\[\frac{15}{28} \]
\[ \frac{5}{28}\]
Solution
\[\text{ We have } , \]
\[\text{ The number of red balls = 5 and } \]
\[\text{ The number of blue balls = 3} \]
\[\text{ Let R be the event of getting a red ball and} \]
\[\text{ B be the event of getting a blue ball .} \]
\[\text{ Now } , \]
\[P\left( \text{ Getting exactly two red balls of the three balls, the first ball being red } \right) = P\left( RB|R \right) + P\left( BR|R \right)\]
\[ = \frac{4}{7} \times \frac{3}{6} + \frac{3}{7} \times \frac{4}{6}\]
\[ = \frac{2}{7} + \frac{2}{7}\]
\[ = \frac{4}{7}\]
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