Chapter 15.10, Problem 6WE

(a)

To determine

To find:The sample space for the experiment.

Answer to Problem 6WE

The sample space for the experiment is $\left\{\begin{array}{l}\left(1,H\right),\left(2,H\right),\left(3,H\right),\left(4,H\right),\left(5,H\right),\left(6,H\right),\\ \left(1,T\right),\left(2,T\right),\left(3,T\right),\left(4,T\right),\left(5,T\right),\left(6,T\right)\end{array}\right\}$ .

Explanation of Solution

Given information:

The given information is the event $A$ that the die shows $5$ or $6$ ,

The event $B$ is the coin shows heads.

Calculation:

An experiment consists of tossing a die and a coin. Let A be the event that the die shows $5$ or $6$ and let $B$ be the event that the coin shows heads.

The sample space for this experiment consists of ordered pairs, in which the first element corresponds to the outcome of the die and the second outcome corresponds to the outcome of the coin. The possible outcomes when a die is tossed are $1,2,3,4,5,$ and $6$ ; the possible outcomes when a coin is tossed are: H, T. Thus the sample space of the experiment can bespecified as follows:

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

Therefore, the sample space for the experiment

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

(b)

To determine

To find: The sample events in $A$ , $B$ , $A\cup B$ and $A\cap B$ .

Answer to Problem 6WE

The sample events in

  $A=\left\{\left(5,H\right),\left(6,H\right),\left(5,T\right),\left(6,T\right)\right\}$ , $B=\left\{\left(1,H\right),\left(2,H\right),\left(3,H\right),\left(4,H\right),\left(5,H\right),\left(6,H\right)\right\}$ , $\left(A\cup B\right)=\left\{\left(1,H\right),\left(2,H\right),\left(3,H\right),\left(4,H\right),\left(5,H\right),\left(6,H\right),\left(5,T\right),\left(6,T\right)\right\}$ and $\left(A\cap B\right)=\left\{\left(5,H\right),\left(6,H\right)\right\}$ .

Explanation of Solution

Given information:

The given information is the event $A$ that the die shows $5$ or $6$ ,

The event $B$ is the coin shows heads.

Calculation:

Let A be the event that the die shows a 5 or 6, and let 8 be the event that the coin shows heads.

The simple event $A$ can thus be specified as:

  $\left\{\left(5,H\right),\left(6,H\right),\left(5,T\right),\left(6,T\right)\right\}$

The simple event $B$ can be specified as:

  $\left\{\left(1,H\right),\left(2,H\right),\left(3,H\right),\left(4,H\right),\left(5,H\right),\left(6,H\right)\right\}$

Therefore, the event $A\cup B$ the union of these two simple events can be specified as:

  $\left\{\left(1,H\right),\left(2,H\right),\left(3,H\right),\left(4,H\right),\left(5,H\right),\left(6,H\right),\left(5,T\right),\left(6,T\right)\right\}$

The event the intersection of the two simple events $A\cap B$ , can be specified as:

  $\left\{\left(5,H\right),\left(6,H\right)\right\}$

Therefore, the sample events in

  $A=\left\{\left(5,H\right),\left(6,H\right),\left(5,T\right),\left(6,T\right)\right\}$ , $B=\left\{\left(1,H\right),\left(2,H\right),\left(3,H\right),\left(4,H\right),\left(5,H\right),\left(6,H\right)\right\}$ , $\left(A\cup B\right)=\left\{\left(1,H\right),\left(2,H\right),\left(3,H\right),\left(4,H\right),\left(5,H\right),\left(6,H\right),\left(5,T\right),\left(6,T\right)\right\}$ and $\left(A\cap B\right)=\left\{\left(5,H\right),\left(6,H\right)\right\}$ .

(c)

To determine

To find: The probability of $A$ , $B$ , $A\cup B$ and $A\cap B$ .

Answer to Problem 6WE

The probability of $P\left(A\right)=\frac{1}{3},P\left(B\right)=\frac{1}{2},P\left(A\cup B\right)=\frac{2}{3}$ , and $P\left(A\cap B\right)=\frac{1}{6}$ .

Explanation of Solution

Given information:

The given information is the event $A$ that the die shows $5$ or $6$ ,

The event $B$ is the coin shows heads.

Calculation:

The total number of outcomes in the sample space for this experiment of tossing a die and a coin consists of $12$ elements.

The simple event A consists of 4 of these outcomes as shown in the previous step. Hence the probability of the event A is given by:

  $\begin{array}{l}P\left(A\right)=\frac{4}{12}\\ P\left(A\right)=\frac{1}{3}\end{array}$

The simple event B consists of 6 outcomes as shown in the previous step Hence the probability of the event is given by:

  $\begin{array}{l}P\left(B\right)=\frac{6}{12}\\ P\left(B\right)=\frac{1}{2}\end{array}$

The event $A\cup B$ consists of 8 outcomes as shown in the previous step. Hence the probability of the event $A\cup B$ is given by:

  $\begin{array}{l}P\left(A\cup B\right)=\frac{8}{12}\\ P\left(A\cup B\right)=\frac{2}{3}\end{array}$

The event in consists of 2 outcomes as shown in the previousstep. Hence the probability of the event $A\cap B$ is given by:

  $\begin{array}{l}P\left(A\cap B\right)=\frac{2}{12}\\ P\left(A\cap B\right)=\frac{1}{6}\end{array}$

Therefore, the probability of $P\left(A\right)=\frac{1}{3},P\left(B\right)=\frac{1}{2},P\left(A\cup B\right)=\frac{2}{3}$ ,and $P\left(A\cap B\right)=\frac{1}{6}$ .

(d)

To determine

To find: The $A$ and $B$ mutually exclusive. Are they independent?

Answer to Problem 6WE

The events $A$ and $B$ are independent and $A$ and $B$ are not mutually exclusive.

Explanation of Solution

Given information:

The given information is the event $A$ that the die shows $5$ or $6$ ,

The event $B$ is the coin shows heads.

Calculation:

Two elements are mutually exclusive if they have no elements in common; that is if their intersection is empty. But the event $A\cap B$ can be expected as $\left(A\cap B\right)=\left\{\left(5,H\right),\left(6,H\right)\right\}$

Hence, $A$ and $B$ are not mutually exclusive. Two events $A$ and $B$ are independent if and only if

  $P\left(A\cap B\right)=P\left(A\right).P\left(B\right)$

  $P\left(A\right)=\frac{1}{3}$

  $P\left(B\right)=\frac{1}{2}$

  $P\left(A\cap B\right)=\frac{1}{6}$

  $\begin{array}{l}P\left(A\cap B\right)=P\left(A\right).P\left(B\right)\\ \frac{1}{6}=\frac{1}{3}\times \frac{1}{2}\end{array}$

Therefore the events $A$ and $B$ are independent. That is the event that the die shows a 5 or 6,and the event that the coin shows heads are independent since the tossing of the coin is done independently of tossing the die.

Therefore, the events $A$ and $B$ are independent and $A$ and $B$ are not mutually exclusive.