Chapter 5.2, Problem 35E

(a)

To determine

Discuss the probability model as a valid probability model.

Answer to Problem 35E

The probability model is valid.

Explanation of Solution

Given information:

Probability model for the prizes:

  

Criteria for valid probability model:

• For each outcome, sum of probabilities should be equal to 1.
• All probabilities should be between 0 and 1 (including both).

Note that

All the probabilities shown in the table are between 0 and 1.

Thus,

First requirement has been satisfied.

Now,

Sum of the probabilities shown in the table:

  $\begin{array}{l}\text{Sum}\text{of}\text{probabilities}=P\left(P\right)+P\left(C\right)+P\left(S\right)+P\left(Hp\right)+P\left(Eh\right)\\ =0.40+0.25+0.15+0.15+0.05=1\end{array}$

The sum of probabilities shown in the table is equal to 1.

Thus,

Second requirement has also been satisfied.

We can say

This is a valid probability model.

(b)

To determine

Probability for a student does not win extra homework.

Answer to Problem 35E

Probability,

  $P\left(E{h}^c\right)=0.95$

Explanation of Solution

Given information:

Probability model for the prizes:

  

According to complement rule,

  $P\left(A^c\right)=P\left(\text{not}A\right)=1-P\left(A\right)$

Note that

The probability of winning extra home work:

  $P\left(Eh\right)=0.05$

Apply complement rule:

  $P\left(E{h}^c\right)=P\left(\text{not}Eh\right)=1-P\left(Eh\right)=1-0.05=0.95$

Thus,

Probability that a student does not win extra homework is 0.95.

(c)

To determine

Probability for a student wins candy or a homework pass.

Answer to Problem 35E

Probability,

  $P\left(C\cup Hp\right)=0.40$

Explanation of Solution

Given information:

Probability model for the prizes:

  

If the two events cannot occur at the same time, then two events are mutually exclusive or disjoint.

According to addition rule for mutually exclusive or disjoint events:

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

Note that

Probability of winning Candy,

  $P\left(C\right)=0.25$

Probability of winning a homework pass,

  $P\left(Hp\right)=0.15$

Since both events are mutually exclusive events, it is impossible to win both candy and a homework pass.

Apply the addition rule for disjoint or mutually exclusive events:

  $\begin{array}{l}P\left(C\cup Hp\right)=P\left(C\text{or}Hp\right)=P\left(C\right)+P\left(Hp\right)\\ =0.25+0.15\\ =0.40\end{array}$

Thus,

Probability that a student wins candy or a homework pass is 0.40.