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

पर्याय

• \[ \frac{15}{29}\]

• \[\frac{15}{56} \]

• \[ \frac{45}{196} \]

• \[ \frac{135}{392}\]

Answer 1

\[\text{ We have } , \]

\[\text{ The number of red balls = 5 and } \]

\[\text{ The number of blue balls = 3} \]

\[\text{ Let R be the event of getting a red ball and} \]

\[ \text{ B be the event of getting a blue ball .}  \]

\[\text{ Now } , \]

\[P\left( \text{ Getting exactly one red ball }  \right) = P\left( RBB \right) + P\left( BRB \right) + P\left( BBR \right)\]

\[ = P\left( R \right) \times P\left( B|R \right) \times P\left( B|RB \right) + P\left( B \right) \times P\left( R|B \right) \times P\left( B|BR \right) + P\left( B \right) \times P\left( B|B \right) \times P\left( R|BB \right)\]

\[ = \frac{5}{8} \times \frac{3}{7} \times \frac{2}{6} + \frac{3}{8} \times \frac{5}{7} \times \frac{2}{6} + \frac{3}{8} \times \frac{2}{7} \times \frac{5}{6}\]

\[ = \frac{5}{56} + \frac{5}{56} + \frac{5}{56}\]

\[ = \frac{15}{56}\]