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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 of getting exactly one red ball is
Options
\[ \frac{15}{29}\]
\[\frac{15}{56} \]
\[ \frac{45}{196} \]
\[ \frac{135}{392}\]
Solution
\[\text{ We have } , \]
\[\text{ The number of red balls = 5 and } \]
\[\text{ The number of blue balls = 3} \]
\[\text{ Let R be the event of getting a red ball and} \]
\[ \text{ B be the event of getting a blue ball .} \]
\[\text{ Now } , \]
\[P\left( \text{ Getting exactly one red ball } \right) = P\left( RBB \right) + P\left( BRB \right) + P\left( BBR \right)\]
\[ = P\left( R \right) \times P\left( B|R \right) \times P\left( B|RB \right) + P\left( B \right) \times P\left( R|B \right) \times P\left( B|BR \right) + P\left( B \right) \times P\left( B|B \right) \times P\left( R|BB \right)\]
\[ = \frac{5}{8} \times \frac{3}{7} \times \frac{2}{6} + \frac{3}{8} \times \frac{5}{7} \times \frac{2}{6} + \frac{3}{8} \times \frac{2}{7} \times \frac{5}{6}\]
\[ = \frac{5}{56} + \frac{5}{56} + \frac{5}{56}\]
\[ = \frac{15}{56}\]
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