Chapter 3.1, Problem 12E

(a)

To determine

Obtain the probability that a student chosen at random does not miss any days of school due to sickness this year.

Answer to Problem 12E

The probability that a student chosen at random does not miss any days of school due to sickness this year is 0.32.

Explanation of Solution

Calculation:

Based on the given information, the following information is known:

$\begin{array}{l}P\left(\begin{array}{l}\text{Students miss exactly }\\ \text{one day of school}\end{array}\right)=0.25\left(=\frac{25}{100}\right)\\ P\left(\text{Miss 2 days}\right)=0.15\left(=\frac{15}{100}\right)\\ P\left(\begin{array}{l}\text{Miss 3 or more days }\\ \text{due to sickness}\end{array}\right)=0.28\left(=\frac{28}{100}\right)\end{array}$

The probability that a student chosen at random does not miss any days of school due to sickness this year is given below:

$\begin{array}{c}P\left(\begin{array}{l}\text{Does not miss any days of }\\ \text{school due to sickness}\end{array}\right)=1-\left\{\begin{array}{l}P\left(\begin{array}{l}\text{Students miss exactly }\\ \text{one day of school}\end{array}\right)+\\ P\left(\text{Miss 2 days}\right)+\\ P\left(\begin{array}{l}\text{Miss 3 or more days }\\ \text{due to sickness}\end{array}\right)\end{array}\right\}\\ =1-\left\{0.25+0.15+0.28\right\}\\ =1-0.68\\ =0.32\end{array}$

(b)

To determine

Obtain the probability that a student chosen at random misses no more than one day.

Answer to Problem 12E

The probability that a student chosen at random misses no more than one day is 0.57.

Explanation of Solution

Calculation:

From Part (a), it known that

$\begin{array}{l}P\left(\begin{array}{l}\text{Does not miss any }\\ \text{days of the school}\end{array}\right)=0.32\\ P\left(\begin{array}{l}\text{Students miss exactly }\\ \text{one day of school}\end{array}\right)=0.25\end{array}$

The probability that a student chosen at random misses no more than one day is given below:

$\begin{array}{c}P\left(\begin{array}{l}\text{Misses no more }\\ \text{than one day}\end{array}\right)=P\left(\begin{array}{l}\text{Does not miss any }\\ \text{days of the school}\end{array}\right)+P\left(\begin{array}{l}\text{Miss exactly one }\\ \text{day of the school}\end{array}\right)\\ =0.32+0.25\\ =0.57\end{array}$

(c)

To determine

Obtain the probability that a student chosen at random misses at least one day.

Answer to Problem 12E

The probability that a student chosen at random misses at least one day is 0.57.

Explanation of Solution

Calculation:

The probability that a student chosen at random misses at least one day is given below:

$\begin{array}{c}P\left(\text{At least one day}\right)=\left\{\begin{array}{l}P\left(\begin{array}{l}\text{Students miss exactly }\\ \text{one day of school}\end{array}\right)+\\ P\left(\text{Miss 2 days}\right)+\\ P\left(\begin{array}{l}\text{Miss 3 or more days }\\ \text{due to sickness}\end{array}\right)\end{array}\right\}\\ =0.25+0.15+0.28\\ =0.68\end{array}$

(d)

To determine

Obtain the probability that neither kid will miss any school, if a parent has two kids at the County D elementary school.

Answer to Problem 12E

The probability that neither kid will miss any school is 0.1024.

Explanation of Solution

Calculation:

In this context, it is assumed that neither kid missing the school is independent.

From Part (a), it is known that $P\left(\begin{array}{l}\text{Does not miss any }\\ \text{days of the school}\end{array}\right)=0.32$

If a parent has two kids at the County D elementary school, the probability that neither kid will miss any school is shown below:

$\begin{array}{c}P\left(\text{Neither kid miss any school}\right)=\left\{\begin{array}{l}P\left({\text{1}}^{st}\text{ will not miss the school}\right)\times \\ P\left(2^{nd}\text{ will not miss the school}\right)\end{array}\right\}\\ =0.32\times 0.32\\ =0.1024\end{array}$

(e)

To determine

Obtain the probability that both kids will miss some school if a parent has two kids at the County D elementary school.

Answer to Problem 12E

The probability that both kids will miss some school is 0.4624.

Explanation of Solution

Calculation:

In this context, it is assumed that each kid missing the school is independent.

From Part (c), it is known that $P\left(\begin{array}{l}\text{miss some days of the school}\\ \text{due to sickness}\end{array}\right)=0.68$

If a parent has two kids at the County D elementary school, the probability that both kids will miss any school is shown below:

$\begin{array}{c}P\left(\text{Both kid miss any school}\right)=\left\{\begin{array}{l}P\left({\text{1}}^{st}\text{ will miss the school}\right)\times \\ P\left(2^{nd}\text{ will miss the school}\right)\end{array}\right\}\\ =0.68\times 0.68\\ =0.4624\end{array}$

(f)

To determine

State whether it is reasonable to make assumption in Part (d) and Part (e).

Explanation of Solution

It is not reasonable to make assumptions in Part (d) and Part (e) because the 1^st kid missing the school due to sickness may affect the other kid going to school, as it increases the probability of the 2^nd kid falling sick.