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An urn contains 9 balls, 2 of which are white, 3 blue and 4 black. 3 balls are drawn at random from the urn. The chance that 2 balls will be of the same colour and the third of a different colour is
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The correct answer is:
$\frac{55}{84}$
Let $E_{1}, E_{2}$ and $E_{3}$ be the the events such that drawn ball is white, blue and black respectively. The required probability
$$
\begin{aligned}
&={ }^{3} P_{1} P\left(E_{1}\right) P\left(E_{1}\right) P(\bar{E})_{1}+{ }^{3} P_{1} P\left(E_{2}\right) P\left(E_{2}\right) P\left(\bar{E}_{2}\right) \\
&+{ }^{3} P_{1} P\left(E_{3}\right) P\left(E_{3}\right) P\left(\bar{E}_{3}\right) \\
&=3 \times \frac{2}{9} \times \frac{1}{8} \times \frac{7}{7}+3 \times \frac{3}{9} \times \frac{2}{8} \times \frac{6}{7}+3 \times \frac{4}{9} \times \frac{3}{8} \times \frac{5}{7} \\
&=3\left[\frac{14+36+60}{9 \times 8 \times 7}\right]=\frac{3 \times 110}{9 \times 8 \times 7}=\frac{55}{84}
\end{aligned}
$$
$$
\begin{aligned}
&={ }^{3} P_{1} P\left(E_{1}\right) P\left(E_{1}\right) P(\bar{E})_{1}+{ }^{3} P_{1} P\left(E_{2}\right) P\left(E_{2}\right) P\left(\bar{E}_{2}\right) \\
&+{ }^{3} P_{1} P\left(E_{3}\right) P\left(E_{3}\right) P\left(\bar{E}_{3}\right) \\
&=3 \times \frac{2}{9} \times \frac{1}{8} \times \frac{7}{7}+3 \times \frac{3}{9} \times \frac{2}{8} \times \frac{6}{7}+3 \times \frac{4}{9} \times \frac{3}{8} \times \frac{5}{7} \\
&=3\left[\frac{14+36+60}{9 \times 8 \times 7}\right]=\frac{3 \times 110}{9 \times 8 \times 7}=\frac{55}{84}
\end{aligned}
$$
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