Chapter 13.5, Problem 19E

To determine

To calculate: The probability that numbers are odd provided that sum of the numbers are even.

Answer to Problem 19E

The probability that numbers are odd provided that sum of the numbers are even is $\frac{5}{8}$ .

Explanation of Solution

Given information:

From a box two tiles numbered from 1 to 9 are drawn at random.

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 from a box two tiles numbered from 1 to 9 are drawn at random.

Let 2 tiles be $\boxed{}\boxed{}$ .

The total number of ways to draw 2 tiles are $9\cdot 8=72$ ways since the tiles are not replaced.

From 1 to 9 there 5 odd numbers and 4 even numbers.

Let A be an event that number is odd and B be an event that sum is even.

The possible combinations such that numbers are odd.

There are 5 ways to choose first tile and 4 ways to choose the second tile as the event is without replacement.

Total number of ways are $5\cdot 4=20$ .

The combinations that sum is even.

Sum of two odd numbers is even and sum of two even numbers is also even.

There are 20 ways to select two odd number.

For sum is even, there are 4 ways to choose first tile and 3 ways to choose the second tile as the event is without replacement.

The number of ways are $4\cdot 3=12$ .

Total number of ways are $20+12=32$ .

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 sum is even.

  $P\left(B\right)=\frac{32}{72}$

Probability that numbers are odd and sum is even.

  $P\left(A\text{ and B}\right)=\frac{20}{72}$

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 numbers are odd provided that sum of the numbers are even.

  $\begin{array}{c}P\left(A|B\right)=\frac{\frac{20}{72}}{\frac{32}{72}}\\ =\frac{20}{32}\\ =\frac{5}{8}\end{array}$

Thus, the probability that numbers are odd provided that sum of the numbers are even is $\frac{5}{8}$ .