Question

A box contains coupons labelled $1,2,\dots ,100$. Five coupons are picked at random one after another without replacement. Let the numbers on the coupons be $x_1,x_2,\dots ,x_5$. What is the probability that $x_1>x_2>x_3$ and $x_3<x_4<x_5$?

• A. $1/120$
• B. $1/60$
• C. $1/20$
• D. $1/10$

Answer 1

The correct option is C $1/20$

first we make the denominator by $100C_5\ast 5!$ the total number of cases

then we need to take favourable number of cases so $x_3$ is always the smallest of all hence it is always fixed now we choose any 2 numbers from rest 4 i.e $4C_2$ ways and assign them as $x_{1}and{x}_2$ so now there is only one way as condition $x_1>x_2$ hence rest 2 numbers are also fixed similarly so the total number of favourable outcomes are $100C_5\ast 4C_2$ hence the answer is $4C_2/5!=1/20$