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A box contains coupons labelled $1,2, \ldots, 100$. Five coupons are picked at random one after another without replacement. Let the numbers on the coupons be $x_1, x_2, \ldots, x_5$. What is the probability that $x_1 > x_2 > x_3$ and $x _3 < x _4 < x _5 ?$
$1 / 120$
$1 / 60$
$1 / 20$
$1 / 10$
Solution
(c)
We have,$100$ coupons labelled $1$ , $2,3, \ldots 100$
Five coupons are random selected and arranged.
$\therefore$ Total numbers of outcomes $={ }^{110} C_5 \times 5$ !
Five coupons $x_1, x_2, x_3, x_4, x_5$ are arranged such that $x_1 > x_2 > x_3$ and
$x_3 < x_4 < x_5$
Favourable outcomes $={ }^{100} C_5 \times \frac{4 !}{2 ! 2 !}$
$\therefore \text { Required probability } =\frac{{ }^{100} C_5 \times \frac{4 !}{2 ! 2 !}}{{ }^{100} C_5 \times 5 !}$
$=\frac{1}{20}$
