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

$\left(\mathrm{a}\right)\frac{15}{29}\left(\mathrm{b}\right)\frac{15}{56}\left(\mathrm{c}\right)\frac{45}{196}\left(\mathrm{d}\right)\frac{135}{392}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{The}\mathrm{number}\mathrm{of}\mathrm{red}\mathrm{balls}=5\mathrm{and} \\ \mathrm{The}\mathrm{number}\mathrm{of}\mathrm{blue}\mathrm{balls}=3 \\ \mathrm{Let}R\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{red}\mathrm{ball}\mathrm{and} \\ B\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{blue}\mathrm{ball}. \\ \mathrm{Now}, \\ \mathrm{P}\left(\mathrm{Getting}\mathrm{exactly}\mathrm{one}\mathrm{red}\mathrm{ball}\right)=\mathrm{P}\left(RBB\right)+\mathrm{P}\left(BRB\right)+\mathrm{P}\left(BBR\right) \\ =\mathrm{P}\left(R\right)\times \mathrm{P}\left(B|R\right)\times \mathrm{P}\left(B|RB\right)+\mathrm{P}\left(B\right)\times \mathrm{P}\left(R|B\right)\times \mathrm{P}\left(B|BR\right)+\mathrm{P}\left(B\right)\times \mathrm{P}\left(B|B\right)\times \mathrm{P}\left(R|BB\right) \\ =\frac{5}{8}\times \frac{3}{7}\times \frac{2}{6}+\frac{3}{8}\times \frac{5}{7}\times \frac{2}{6}+\frac{3}{8}\times \frac{2}{7}\times \frac{5}{6} \\ =\frac{5}{56}+\frac{5}{56}+\frac{5}{56} \\ =\frac{15}{56} \end{gathered}$$

Hence, the correct alternative is option (b).