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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) $\frac{14}{55}$Solution
A bag contains 10 balls out of which k are red and ( $10-\mathrm{k}$ ) are black
There are 8 possibilities as
$\mathrm{K}=0\left(\mathrm{E}_{0}\right) 0$ Red 10 Black
$\mathrm{K}=1\left(\mathrm{E}_{1}\right) \quad 1$ Red 9 Black
$\mathrm{K}=2\left(\mathrm{E}_{2}\right) 2$ Red 8 Black
$\mathrm{K}=3\left(\mathrm{E}_{3}\right) 3$ Red 7 Black
$\mathrm{K}=7\left(\mathrm{E}_{7}\right) 7$ Red 3 Black
$\mathrm{P}\left(\mathrm{E}_{1}\right)=\mathrm{P}\left(\mathrm{E}_{2}\right)=\mathrm{P}\left(\mathrm{E}_{3}\right)=\ldots=\mathrm{P}\left(\mathrm{E}_{7}\right)=\mathrm{p}$
$\mathrm{P}\binom{\text { all 3 }}{\text { are black }}=\mathrm{p} \frac{{ }^{10} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}}+\mathrm{p} \frac{{ }^{9} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}}+\mathrm{p} \frac{{ }^{8} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}}+\cdots+\mathrm{p} \frac{{ }^{3} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}}$
$\mathrm{P}\left(\mathrm{E}_{1} /\right.$ All3Blacks $)=\frac{{ }^{9} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}+{ }^{9} \mathrm{C}_{3}+{ }^{8} \mathrm{C}_{3}+\ldots .+{ }^{3} \mathrm{C}_{3}}$
$\left(\because{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}={ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}+1}\right)$
${ }^{3} \mathrm{C}_{3}+{ }^{4} \mathrm{C}_{3}+{ }^{5} \mathrm{C}_{3}+\cdots+{ }^{8} \mathrm{C}_{3}+{ }^{9} \mathrm{C}_{3}+{ }^{10} \mathrm{C}_{3}$
$={ }^{4} \mathrm{C}_{4}+{ }^{4} \mathrm{C}_{3}+{ }^{5} \mathrm{C}_{3}+\ldots .+{ }^{9} \mathrm{C}_{3}+{ }^{10} \mathrm{C}_{3}$
$={ }^{5} \mathrm{C}_{4}+{ }^{5} \mathrm{C}_{3}+\ldots .+{ }^{9} \mathrm{C}_{3}+{ }^{10} \mathrm{C}_{3}$
$={ }^{6} \mathrm{C}_{4}+\ldots .+{ }^{9} \mathrm{C}_{3}+{ }^{10} \mathrm{C}_{3}$
$={ }^{10} \mathrm{C}_{4}+{ }^{10} \mathrm{C}_{3}={ }^{11} \mathrm{C}_{4}$
Pr ob. $=\frac{{ }^{9} \mathrm{C}_{3}}{{ }^{11} \mathrm{C}_{4}}=\frac{9.8 .7}{6\left(\frac{11.10 .9 .8}{24}\right)}=\frac{14}{55}$ Answer(4)
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