Chapter 5.2, Problem 60E

To determine

To find:

The event $7or11$ and the probability of $P\left(7or11\right)$

Answer to Problem 60E

The probability of $P\left(7or11\right)$

  $=\frac{8}{36}$

Explanation of Solution

Given:

The table

Outcome	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
Probability	$1/36$	$2/36$	$3/36$	$4/36$	$5/36$	$6/36$	$5/36$	$4/36$	$3/36$	$2/36$	$1/36$

  $\begin{array}{l}P\left(AorB\right)=P\left(A\cup B\right)\\ P\left(AorB\right)=P\left(A\right)+P\left(B\right)\end{array}$

Concept used:

Definition of addition theorem of probability:- let A and B are two events of a random experiment and interested to find out the probability A or B

Then use the addition theorem of probability

Calculation:

let A and B are two events of a random experiment and interested to find out the probability A or B

Then use the addition theorem of probability

That is $\begin{array}{l}P\left(AorB\right)=P\left(A\cup B\right)\\ P\left(AorB\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)\end{array}$

If A and B are mutually exclusive events

Then $P\left(A\cap B\right)=0$

  $\begin{array}{l}P\left(7or11\right)=P\left(A\right)+P\left(B\right)\\ P\left(7or11\right)=\frac{8}{36}\end{array}$

Therefore,

The event $AorB$ and the probability of $P\left(7or11\right)$

  $=\frac{8}{36}$