Chapter 13, Problem 10PT

To determine

Find the probability that a number using the numbers $7,7,7,2,2,2\text{ and }6$ will be $622-2777$ .

Answer to Problem 10PT

The probability of the number $622-2777$ is $\frac{1}{140}$ .

Explanation of Solution

Given information: The given numbers are $7,7,7,2,2,2\text{ and }6$ .

Calculation: The number comprises of 7 digits in which 2 occurs twice and 7 occurs thrice. Therefore the number of distinguish permutations of 7-digit number are $\frac{7!}{3!3!}$ .

Simplify $\frac{7!}{3!3!}$ gives $\frac{7!}{3!3!}=140$ . Now there is only 1 way for which the outcome can be $622-2777$ among 140 cases. Therefore the probability $P\left(622-2777\right)$ is evaluated as,

  $\begin{array}{c}P\left(622-2777\right)=\frac{\text{Favourable cases}}{\text{Total cases}}\\ =\frac{1}{140}\end{array}$

Hence the probability of the number $622-2777$ is $\frac{1}{140}$ .