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Bag A contains 4 white and 2 black balls, bag B contains
3 white and 3 black balls and bag C contains 2 white and
4 black balls. If a bag is chosen at random and a ball is
chosen at random from it, then the probability that the
ball drawn is black is
Options:
3 white and 3 black balls and bag C contains 2 white and
4 black balls. If a bag is chosen at random and a ball is
chosen at random from it, then the probability that the
ball drawn is black is
Solution:
1213 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{2}$
$\mathrm{P}[$ Ball is black $]=\mathrm{P}\left[1^{\text {st }}\right.$ bag selected $] \times$
$$
\mathrm{P}\left[\frac{\text { Ball is black }}{1^{\text {st }} \text { bag selected }}\right]
$$
$$
\begin{gathered}
+\mathrm{P}\left[\mathrm{II}^{\text {nd }} \text { bag selected }\right] \times \mathrm{P}\left[\frac{\text { Ball is black }}{\mathrm{II}^{\text {nd }} \text { bag selected }}\right]+ \\
\mathrm{P}\left[\mathrm{III}^{\text {rd }} \text { bag selected }\right] \times \mathrm{P}\left[\frac{\text { Ball is blank }}{\mathrm{III}^{\text {rd }} \text { bag selected }}\right]
\end{gathered}
$$
$\Rightarrow \mathrm{P}[$ Ball is black $]=\mathrm{P}\left[\right.$ Ball is black $\Lambda 1^{\text {st }}$ bag selected $]+$ $\mathrm{P}\left[\right.$ Ball is black $\Lambda$ II ${ }^{\text {nd }}$ bag selected $]+\mathrm{P}[$ Ball is black $\Lambda$ III ${ }^{\text {rd }}$ bag selected]
$$
\begin{aligned}
& \Rightarrow \mathrm{P}[\text { Ball is black }]=\frac{2}{6} \times \frac{1}{3}+\frac{3}{6} \times \frac{1}{3}+\frac{4}{6} \times \frac{1}{3} \\
& \Rightarrow \mathrm{P}[\text { Ball is black }]=\frac{2}{18}+\frac{3}{18}+\frac{4}{18} \frac{9}{18}=\frac{1}{2}
\end{aligned}
$$
$$
\mathrm{P}\left[\frac{\text { Ball is black }}{1^{\text {st }} \text { bag selected }}\right]
$$
$$
\begin{gathered}
+\mathrm{P}\left[\mathrm{II}^{\text {nd }} \text { bag selected }\right] \times \mathrm{P}\left[\frac{\text { Ball is black }}{\mathrm{II}^{\text {nd }} \text { bag selected }}\right]+ \\
\mathrm{P}\left[\mathrm{III}^{\text {rd }} \text { bag selected }\right] \times \mathrm{P}\left[\frac{\text { Ball is blank }}{\mathrm{III}^{\text {rd }} \text { bag selected }}\right]
\end{gathered}
$$
$\Rightarrow \mathrm{P}[$ Ball is black $]=\mathrm{P}\left[\right.$ Ball is black $\Lambda 1^{\text {st }}$ bag selected $]+$ $\mathrm{P}\left[\right.$ Ball is black $\Lambda$ II ${ }^{\text {nd }}$ bag selected $]+\mathrm{P}[$ Ball is black $\Lambda$ III ${ }^{\text {rd }}$ bag selected]
$$
\begin{aligned}
& \Rightarrow \mathrm{P}[\text { Ball is black }]=\frac{2}{6} \times \frac{1}{3}+\frac{3}{6} \times \frac{1}{3}+\frac{4}{6} \times \frac{1}{3} \\
& \Rightarrow \mathrm{P}[\text { Ball is black }]=\frac{2}{18}+\frac{3}{18}+\frac{4}{18} \frac{9}{18}=\frac{1}{2}
\end{aligned}
$$
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