Question

Three numbers are to be selected at random without replacement from the set of numbers {1, 2, ... n}. The conditional probability that the third number lies between the first two if the first number is known to be smaller than the second is :

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{5}{6}$
• D. $\frac{7}{12}$

Answer 1

The correct option is A $\frac{1}{3}$

Consider the following events
A = The first number is less than the second number
B = The third number lies between the first and the second.
Now, we have to find $P\left(\frac{B}{A}\right)$
Also, we have $P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P\left(A\right)}$
Any 3 numbers can be chosen out of n numbers in ${}^n{C}_3$ ways.
Let the selected numbers be $x_1.x_2.x_3$. Then they satisfy exactly one of the following inequalities.
$x_1<x_2<x_3,x_1<x_3<x_2,x_2<x_1<x_3,x_2<x_3<x_1,x_3<x_1<x_2,x_3<x_2<x_1$
The total number of ways of selecting three numbers and then arranging them = ${}^n{C}_3\times 3!{=}^n{P}_3$
$\therefore P\left(A\right)=\frac{{}^n{C}_3\times 3}{{}^n{C}_3\times 3!}$
and $P(A\cap B)\frac{{}^n{C}_3}{{}^n{C}_3\times 3!}$
Hence $P\left(\frac{B}{A}\right)\frac{P(A\cap B)}{P\left(A\right)}=\frac{1}{3}$