$2$ bad oranges are accidentally mixed with $10$ good ones. $3$ oranges are drawn at random without replacement from this lot. Obtain the probability distribution for the number of bad oranges.
$3$ oranges can be drawn, without replacement, from $12$ oranges in ${^{12}}{P}_{3} = \dfrac{12!}{9!}$ ways
Let $X$ denote a random variable which is drawing bad oranges when $3$ oranges are selected without replacement.
$X$ can take values $0, \; 1, \; 2$
Now,
$P \left(X = 0\right) = P \left(\text{No bad orange is selected}\right)$
$ = \dfrac{{^{10}}{P}_{3}}{{^{12}}{P}_{3}} = \dfrac{\dfrac{10!}{7!}}{\dfrac{12!}{9!}} = \dfrac{10! \times 9!}{12! \times 7!} = \dfrac{9 \times 8}{12 \times 11} = \dfrac{6}{11}$
$P \left(X = 1\right) = P \left(1\text{ bad orange is selected}\right)$
$ = \dfrac{3 \times {^{2}}{P}_{1} \times {^{10}}{P}_{2}}{{^{12}}{P}_{3}} = \dfrac{3 \times 2 \times\dfrac{10!}{8!}}{\dfrac{12!}{9!}} = \dfrac{3 \times 2 \times 10! \times 9!}{12! \times 8!} = \dfrac{3 \times 2 \times 9}{12 \times 11} = \dfrac{9}{22}$
$P \left(X = 2\right) = P \left(2\text{ bad oranges are selected}\right)$
$ = \dfrac{3 \times 2 \times {^{10}}{P}_{1}}{{^{12}}{P}_{3}} = \dfrac{3 \times 2 \times\dfrac{10!}{9!}}{\dfrac{12!}{9!}} = \dfrac{3 \times 2 \times 10! }{12!} = \dfrac{3 \times 2}{12 \times 11} = \dfrac{1}{22}$
Hence the probability distribution of the random variable $X$ is given by
| $X$ | $0$ | $1$ | $2$ |
|---|---|---|---|
| $P \left(X\right)$ | $\dfrac{6}{11} = \dfrac{12}{22}$ | $\dfrac{9}{22}$ | $\dfrac{1}{22}$ |