[shahkb4]

A bag contains 10 balls out of which k are red and ( $10-\mathrm{k}$ ) are black, where $0 \leq \mathrm{k} \leq 10$. If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is [JEE MAIN 28 Jan 2026 S I]
(1) $\frac{7}{11}$
(2) $\frac{7}{55}$
(3) $\frac{7}{110}$
(4) $\frac{14}{55}$

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[shahkb4]

Solution

A bag contains 10 balls out of which k are red and (10-k) are black
There are 8 possibilities as

\[ \begin{aligned} & \frac{\mathrm{K}=0\left(\mathrm{E}_{0}\right) 0 \text { Red } 10 \text { Black }}{\mathrm{K}=1\left(\mathrm{E}_{1}\right) 1 \text { Red } 9 \text { Black }} \\ & \hline \mathrm{K}=2\left(\mathrm{E}_{2}\right) 2 \text { Red } 8 \text { Black } \\ & \mathrm{K}=3\left(\mathrm{E}_{3}\right) 3 \text { Red } 7 \text { Black } \\ & \mathrm{K}=7\left(\mathrm{E}_{7}\right) 7 \text { Red } 3 \text { Black } \\ & \mathrm{P}\left(\mathrm{E}_{1}\right)=\mathrm{P}\left(\mathrm{E}_{2}\right)=\mathrm{P}\left(\mathrm{E}_{3}\right)=\ldots=\mathrm{P}\left(\mathrm{E}_{7}\right)=\mathrm{p} \\ & \mathrm{P}\binom{\text { all } 3}{\text { are black }}=\mathrm{p} \frac{{ }^{10} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}}+\mathrm{p} \frac{{ }^{9} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}}+\mathrm{p} \frac{{ }^{8} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}}+\cdots+\mathrm{p} \frac{{ }^{3} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}} \\ & \mathrm{P}\left(\mathrm{E}_{1} / \text { All } 3 \text { Blacks }\right)=\frac{{ }^{9} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}+{ }^{9} \mathrm{C}_{3}+{ }^{8} \mathrm{C}_{3}+\ldots .+{ }^{3} \mathrm{C}_{3}} \\ & \left(\because{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}={ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}+1}\right) \\ & { }^{3} \mathrm{C}_{3}+{ }^{4} \mathrm{C}_{3}+{ }^{5} \mathrm{C}_{3}+\ldots+{ }^{8} \mathrm{C}_{3}+{ }^{9} \mathrm{C}_{3}+{ }^{10} \mathrm{C}_{3} \\ & ={ }^{4} \mathrm{C}_{4}+{ }^{4} \mathrm{C}_{3}+{ }^{5} \mathrm{C}_{3}+\ldots .+{ }^{9} \mathrm{C}_{3}+{ }^{10} \mathrm{C}_{3} \\ & ={ }^{5} \mathrm{C}_{4}+{ }^{5} \mathrm{C}_{3}+\ldots .+{ }^{9} \mathrm{C}_{3}+{ }^{10} \mathrm{C}_{3} \\ & ={ }^{6} \mathrm{C}_{4}+\ldots .+{ }^{9} \mathrm{C}_{3}+{ }^{10} \mathrm{C}_{3} \\ & ={ }^{10} \mathrm{C}_{4}+{ }^{10} \mathrm{C}_{3}={ }^{11} \mathrm{C}_{4} \\ & \operatorname{Pr} \text { ob. }=\frac{{ }^{9} \mathrm{C}_{3}}{{ }^{11} \mathrm{C}_{4}}=\frac{9.8 .7}{6\left(\frac{11.10 .9 .8}{24}\right)}=\frac{14}{55} \text { Answer(4) } \end{aligned} \]

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