Chapter 12, Problem 12.33E

a.

To determine

To identify: Whether the probabilities assigned to the outcomes are valid.

To state: The reason for the probability model to be invalid.

Answer to Problem 12.33E

Answers:

The probabilities assigned to the outcomes are valid because the given probability model satisfies all probability rules.

Explanation of Solution

Given info:

Rolling a six sided die yields the following results:

$\begin{array}{l}P\left(1\right)=0\\ P\left(2\right)=\frac{1}{6}\\ P\left(3\right)=\frac{1}{3}\\ P\left(4\right)=\frac{1}{3}\\ P\left(5\right)=\frac{1}{6}\\ P\left(6\right)=0\end{array}$

Justification:

Probability Rules:

• The sum of the probabilities for all possible outcomes must be equal to 1. That is, for sample space S in the probability model, $P\left(S\right)=1$
• The probability for any event lies between 0 and 1, inclusively. That is, for an event A, $0\le P\left(A\right)\le 1$
• The events A and B are disjoint events if they don’t have any outcomes in common, and because of this fact, the two events cannot occur together.

$P\left(A\text{ or }B\right)=P\left(A\right)+P\left(B\right)$

• For the event A, $P\left(\text{Event }A\text{ not occuring}\right)=1-P\left(A\right)$

The sum of all individual probabilities are calculated as,

$\begin{array}{c}\text{Total Probability = Sum of all Individual Probabilities}\\ \text{=}P\left(1\right)+P\left(2\right)+P\left(3\right)+P\left(4\right)+P\left(5\right)+P\left(6\right)\\ =0+\frac{1}{6}+\frac{1}{3}+\frac{1}{3}+\frac{1}{6}+0\\ =1\end{array}$

Thus, the sum of all individual probabilities is equal to 1.

All individual probabilities lie between 0 and 1.

All the events in the given probability model are disjoint because the events do not have outcomes in common.

Thus, the given probability model is a valid probability model.

b.

To determine

To identify: Whether the probabilities assigned to the outcomes are valid.

To state: The reason for the probability model to be invalid.

Answer to Problem 12.33E

The probabilities assigned to the outcomes are valid because the given probability model satisfies all probability rules.

Explanation of Solution

Given info:

A card is selected from each type in a deck of cards. The probability for each selection is given below:

$\begin{array}{c}P\left(\text{Clubs}\right)=\frac{12}{52}\\ P\left(\text{Diamonds}\right)=\frac{12}{52}\\ P\left(\text{Hearts}\right)=\frac{12}{52}\\ P\left(\text{Spades}\right)=\frac{16}{52}\end{array}$

Justification:

The sum of all individual probabilities are calculated as,

$\begin{array}{c}\text{Total Probability = Sum of all Individual Probabilities}\\ \text{=}P\left(\text{Clubs}\right)+P\left(\text{Diamonds}\right)+P\left(\text{Hearts}\right)+P\left(\text{Spades}\right)\\ =\frac{12}{52}+\frac{12}{52}+\frac{12}{52}+\frac{16}{52}\\ =1\end{array}$

Thus, the sum of all individual probabilities is equal to 1.

All individual probabilities lie between 0 and 1.

All the events in the given probability model are disjoint because the events do not have outcomes in common.

Thus, the given probability model is a valid probability model.

c.

To determine

To identify: Whether the probabilities assigned to the outcomes are valid.

To state: The reason for the probability model to be invalid.

Answer to Problem 12.33E

The probabilities assigned to the outcomes are not valid because the given probability model does not satisfy the rule “The sum of all individual probabilities must be equal to 1”.

Explanation of Solution

Given info:

College students are selected at random and their sex and enrollment statuses are recorded. The probabilities are given below:

$\begin{array}{c}P\left(\text{Female full-time}\right)=0.56\\ P\left(\text{Male full-time}\right)=0.44\\ P\left(\text{Female part-time}\right)=0.24\\ P\left(\text{Male part-time}\right)=0.17\end{array}$

Justification:

The sum of all individual probabilities are calculated as,

$\begin{array}{c}\text{Total Probability}=\text{Sum of all Individual Probabilities}\\ =\left[\begin{array}{c}P\left(\text{Female full-time}\right)+P\left(\text{Male full-time}\right)+P\left(\text{Female part-time}\right)\\ +P\left(\text{Male part-time}\right)\end{array}\right]\\ =0.56+0.44+0.24+0.17\\ =1.41\end{array}$

                             $\ne 1$

Thus, the sum of all individual probabilities is not equal to 1.

All individual probabilities lie between 0 and 1.

All the events in the given probability model are disjoint because the events do not have outcomes in common.

Thus, the given probability model is not a valid probability model as the rule “sum of all individual probabilities is not equal to 1” is violated.