Chapter 13.5, Problem 18E

To determine

To calculate: The probability that number ends in the digits 52 provided that it is even.

Answer to Problem 18E

The probability that number ends in the digits 52 provided that it is even is $\frac{1}{8}$ .

Explanation of Solution

Given information:

A five digit number is to be constructed with digits 1, 2, 3, 4 and 5.

Formula used:

Probability of an event E is $P\left(\text{E}\right)=\frac{\text{number of outcome}}{\text{total number of outcome}}$ .

Given an event Q, the conditional probability of event A is $P\left(A|Q\right)=\frac{P\left(A\text{ and Q}\right)}{P\left(Q\right)}$ where $P\left(Q\right)\ne 0$ .

Calculation:

Consider the provided information that five digit number is to be constructed with digits 1, 2, 3, 4 and 5.

Let 5-digit number be $\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}$ .

The total number of five digit number will be $\text{5!}$ when each digit is used once.

Let A be an event that number is of the form $\boxed{}\boxed{}\boxed{}\boxed{5}\boxed{2}$ and B be an event that number is even.

The possible combinations such that number is even is that either the number will end with 2 or 4.

Case 1: When number ends with 2 that is of the form $\boxed{}\boxed{}\boxed{}\boxed{}\boxed{2}$

There are 4 ways to fill first box, 3 ways to fill the second box, 2 ways to fill the third box and 1 way to fill the fourth box.

There are total $4\cdot 3\cdot 2\cdot 1=24$ ways.

Case 2: When number ends with 4 that is of the form $\boxed{}\boxed{}\boxed{}\boxed{}\boxed{4}$

There are 4 ways to fill first box, 3 ways to fill the second box, 2 ways to fill the third box and 1 way to fill the fourth box.

There are total $4\cdot 3\cdot 2\cdot 1=24$ ways.

Therefore, total number of 5 digit number that are even are $24+24=48$ .

The combinations that number is of the form $\boxed{}\boxed{}\boxed{}\boxed{5}\boxed{2}$ and number is even.

There are 3 ways to fill first box, 2 ways to fill the second box, 1 ways to fill the third box.

There are total $3\cdot 2\cdot 1=6$ ways.

Recall that probability of an event E is $P\left(\text{E}\right)=\frac{\text{number of outcome}}{\text{total number of outcome}}$ .

Probability that event that number is even.

  $P\left(B\right)=\frac{48}{5!}$

Probability that number is of the form $\boxed{}\boxed{}\boxed{}\boxed{5}\boxed{2}$ number is even.

  $P\left(A\text{ and B}\right)=\frac{6}{5!}$

Recall that given an event Q, the conditional probability of event A is $P\left(A|Q\right)=\frac{P\left(A\text{ and Q}\right)}{P\left(Q\right)}$ where $P\left(Q\right)\ne 0$ .

The probability that number ends in the digits 52 provided that it is even.

  $\begin{array}{c}P\left(A|B\right)=\frac{\frac{6}{5!}}{\frac{48}{5!}}\\ =\frac{6}{48}\\ =\frac{1}{8}\end{array}$

Thus, the probability that number ends in the digits 52 provided that it is even is $\frac{1}{8}$ .