Chapter 11.3, Problem 25PPSE

Solution

To identify:The probability $P\left(3orodd\right)$ .A standard number cube is tossed.

The probability is $P\left(3andodd\right)=\frac{1}{2}$

Given information:

A standard cube is tossed.

Explanation:

Consider the information.

Since there are six faces numbered as 1, 2, 3, 4, 5, 6,

So sample space of rolling a cube is, $S=\left\{1,2,3,4,5,6\right\}$

As 3 exist one time on dice, so probability of getting 3 is $P\left(3\right)=\frac{1}{6}$

There are three odd numbers $1,3,5$

So, $P\left(odd\right)=\frac{3}{6}$

As 3 exist in odd numbers too

That means these events are not mutually exclusive

That is. $P\left(3andodd\right)\ne 0$

Here, $P\left(3andodd\right)=\frac{1}{6}$

For non-mutually exclusive events A and B

For non-mutually exclusive events A and B

  $P\left(AorB\right)=P\left(A\right)+P\left(B\right)-P\left(AandB\right)$

Therefore, $P\left(3orodd\right)=P\left(3\right)+P\left(odd\right)-P\left(3andodd\right)\mathrm{...}\left(1\right)$

Substitute the values in equation (1) and solve

  $\begin{array}{c}P\left(3orodd\right)=P\left(3\right)+P\left(odd\right)-P\left(3andodd\right)\\ P\left(3orodd\right)=\frac{1}{6}+\frac{3}{6}-\frac{1}{6}\\ =\frac{3}{6}\\ =\frac{1}{2}\end{array}$

Hence $P\left(3andodd\right)=\frac{1}{2}$