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The following table shows the probability of selecting the boxes $A, B$ and $C$ and the number of balls of different colours contained in them.
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1414 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{13}$
Probability of green ball $P(G)$
$$
\begin{aligned}
& =\frac{1}{2} \times \frac{2}{6}+\frac{1}{3} \times \frac{3}{6}+\frac{1}{6} \times \frac{1}{6}=\frac{1}{6}+\frac{1}{6}+\frac{1}{36} \\
& =\frac{6+6+1}{36}=\frac{13}{36}
\end{aligned}
$$
Let probability of drawn ball is green comes from bag $C$ is $P\left(\frac{C}{G}\right)$, then
$$
P\left(\frac{C}{G}\right)=\frac{P\left(\frac{G}{C}\right) \times P(C)}{P(G)}=\frac{\frac{1}{6} \times \frac{1}{6}}{\frac{13}{36}}=\frac{1}{36} \times \frac{36}{13}=\frac{1}{13}
$$
$$
\begin{aligned}
& =\frac{1}{2} \times \frac{2}{6}+\frac{1}{3} \times \frac{3}{6}+\frac{1}{6} \times \frac{1}{6}=\frac{1}{6}+\frac{1}{6}+\frac{1}{36} \\
& =\frac{6+6+1}{36}=\frac{13}{36}
\end{aligned}
$$
Let probability of drawn ball is green comes from bag $C$ is $P\left(\frac{C}{G}\right)$, then
$$
P\left(\frac{C}{G}\right)=\frac{P\left(\frac{G}{C}\right) \times P(C)}{P(G)}=\frac{\frac{1}{6} \times \frac{1}{6}}{\frac{13}{36}}=\frac{1}{36} \times \frac{36}{13}=\frac{1}{13}
$$
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