Chapter 9.3, Problem 18PSB

To determine

To find: The probability for given lengths $AD,CD,CD\text{ and }BD$ .

Answer to Problem 18PSB

  $\text{The required probability }=\frac{3}{4}$

Explanation of Solution

Given information There are four lengths like $AD,CD,CD\text{ and }BD$ .

There are 4 lengths. And one length is repeated. So, only three lengths are like $AD,CD\text{ and }BD$ .

There are $3!$ ways to arrange these lengths.

  $3!=3\cdot 2=6\text{ ways }$

But we have to make two ratios at a time. So, there will be only 3 ways to form 2 ratios

Therefore, required probability $=\frac{3}{4}$