Chapter 3.1, Problem 19GP

a.

To determine

Explain how to represent event $A^c$.

Answer to Problem 19GP

Event $A^c$ can be represented as the total that is equal to 12.

Explanation of Solution

Calculation:

The possible outcomes when rolling two dice is given below:

$\left\{\begin{array}{l}\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\\ \left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right),\\ \left(3,1\right),\left(3,2\right),\left(3,3\right),\left(4,4\right),\left(3,5\right),\left(3,6\right),\\ \left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(4,6\right),\\ \left(5,1\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,5\right),\left(5,6\right),\\ \left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)\end{array}\right\}$

The sum of the outcomes is given below:

$S=\left\{\begin{array}{l}2,3,4,5,6,7,\\ 3,4,5,6,7,8,\\ 4,5,6,7,8,9,\\ 5,6,7,8,9,10,\\ 6,7,8,9,10,11,\\ 7,8,9,10,11,12\end{array}\right\}$

Complement of an event:

The complement of event W is the event that W does not occur. That is, $W^c$ represents that all the outcomes are not in W.

Event A represents the outcomes of getting a total that is less than 12 when rolling two dice.

Therefore, the complement of A $\left(A^c\right)$ represents that the total is equal to 12.

b.

To determine

Find the probability of $P\left(A^c\right)$.

Answer to Problem 19GP

The probability of $P\left(A^c\right)\text{ is }\frac{1}{36}$.

Explanation of Solution

Calculation:

From Figure 3.5, it can be observed that the total number of favorable outcomes for event $A^c$ is 1.

The total number of outcomes is 36.

The required probability is obtained as follows:

$\begin{array}{c}P\left(A^c\right)=\frac{\text{Favorable number of outcomes for event }{A}^c}{\text{Total number of outcomes}}\\ =\frac{1}{36}\end{array}$

Thus, the probability of $P\left(A^c\right)\text{ is }\frac{1}{36}.$

c.

To determine

Find the probability of $P\left(A\right)$.

Answer to Problem 19GP

The probability of $P\left(A\right)\text{ is }\frac{1}{36}$.

Explanation of Solution

Calculation:

The total number of outcomes is 36.

The required probability is obtained as follows:

$\begin{array}{c}P\left(A\right)=1-P\left(A^c\right)\\ =1-\frac{1}{36}\langle \text{From Part (b)}\rangle \\ =\frac{35}{36}\end{array}$

Thus, the probability of $P\left(A\right)\text{ is }\frac{1}{36}$.