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A bag contains four balls. Two balls are drawn randomly and found them to be white. The probability that all the balls in the bag are white is
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Verified Answer
The correct answer is:
$\frac{3}{5}$
Let the events are
$E$ : Two drawn balls are white; $E_1$ : All balls are white $\mathrm{E}_2: 3$ white balls and 1 other; $\mathrm{E}_3: 2$ balls white and 2 others
$$
\begin{aligned}
& \mathrm{P}\left(\mathrm{E}_1\right)=\frac{1}{3}, \mathrm{P}\left(\mathrm{E}_2\right)=\frac{1}{3}, \mathrm{P}\left(\mathrm{E}_3\right)=\frac{1}{3} \\
& \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{E}_1}\right)=\frac{{ }^4 \mathrm{C}_2}{{ }^4 \mathrm{C}_2}=1 \\
& \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{E}_2}\right)=\frac{{ }^3 \mathrm{C}_2}{{ }^4 \mathrm{C}_2}=\frac{3}{6}=\frac{1}{2} \\
& \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{E}_3}\right)=\frac{{ }^2 \mathrm{C}_2}{{ }^4 \mathrm{C}_2}=\frac{1}{6} \\
& \mathrm{P}\left(\frac{\mathrm{E}_1}{\mathrm{E}}\right)=\frac{\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{E}_1}\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{E}_1}\right)+\mathrm{P}\left(\mathrm{E}_2\right) \cdot \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{E}_2}\right)+\mathrm{P}\left(\mathrm{E}_3\right) \times \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{E}_3}\right)} \\
& =\frac{1 \times \frac{1}{3}}{\left(1+\frac{1}{3}\right)+\left(\frac{1}{2} \times \frac{1}{3}\right)+\left(\frac{1}{6} \times \frac{1}{3}\right)}=\frac{3}{5}
\end{aligned}
$$
$E$ : Two drawn balls are white; $E_1$ : All balls are white $\mathrm{E}_2: 3$ white balls and 1 other; $\mathrm{E}_3: 2$ balls white and 2 others
$$
\begin{aligned}
& \mathrm{P}\left(\mathrm{E}_1\right)=\frac{1}{3}, \mathrm{P}\left(\mathrm{E}_2\right)=\frac{1}{3}, \mathrm{P}\left(\mathrm{E}_3\right)=\frac{1}{3} \\
& \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{E}_1}\right)=\frac{{ }^4 \mathrm{C}_2}{{ }^4 \mathrm{C}_2}=1 \\
& \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{E}_2}\right)=\frac{{ }^3 \mathrm{C}_2}{{ }^4 \mathrm{C}_2}=\frac{3}{6}=\frac{1}{2} \\
& \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{E}_3}\right)=\frac{{ }^2 \mathrm{C}_2}{{ }^4 \mathrm{C}_2}=\frac{1}{6} \\
& \mathrm{P}\left(\frac{\mathrm{E}_1}{\mathrm{E}}\right)=\frac{\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{E}_1}\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{E}_1}\right)+\mathrm{P}\left(\mathrm{E}_2\right) \cdot \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{E}_2}\right)+\mathrm{P}\left(\mathrm{E}_3\right) \times \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{E}_3}\right)} \\
& =\frac{1 \times \frac{1}{3}}{\left(1+\frac{1}{3}\right)+\left(\frac{1}{2} \times \frac{1}{3}\right)+\left(\frac{1}{6} \times \frac{1}{3}\right)}=\frac{3}{5}
\end{aligned}
$$
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