Chapter 4.2, Problem 44E

a.

To determine

Compute the probability that selected report is open and comes from Brooklyn.

Answer to Problem 44E

The probability that selected report is open and comes from Brooklyn is 0.0806.

Explanation of Solution

Calculation:

There were a total of 14,518 reports. The reports are obtained from five Borough. The reports are classified into three categories open, closed and pending. The table provide information about these categories.

The probability of the event A can be obtained by the formula:

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Number of outcomes in}\text{sample space}}$

Here, event A denotes that selected report is open and comes from Brooklyn. Among the 14,518 reports, 1,170 of them are from Brooklyn.

Substitute 1,170 for “number of outcomes in A” and 14,518 for “Number of outcomes in sample space” in the probability formula.

Therefore,

$\begin{array}{c}P\left(A\right)=\frac{1,170}{14,518}\\ =0.08058\\ =0.0806\end{array}$

Thus, the probability that selected report is open and comes from Brooklyn is 0.0806.

b.

To determine

Compute the probability that selected report is closed or comes from Queens.

Answer to Problem 44E

The probability that selected report is closed or comes from Queens is 0.7747.

Explanation of Solution

Calculation:

Event V denotes that selected report is closed, event W denote that selected report is from Queens.

The probability that a randomly selected report is closed or comes from Queens can be expressed as, $P\left(V\text{or }W\right)$.

General Addition rule:

For any two events V and W the general addition rule states that

$P\left(V\text{or }W\right)=P\left(V\right)+P\left(W\right)-P\left(V\text{ and }W\right)$

From table, $P\left(V\right)=\frac{9,869}{14,518}\text{, }P\left(W\right)=\frac{3,421}{14,518},\text{and}P\left(V\text{and }W\right)=\frac{2,043}{14,518}$.

Substitute these values in the general addition rule.

Therefore,

$\begin{array}{c}P\left(V\text{or }W\right)=\frac{9,869}{14,518}+\frac{3,421}{14,518}-\frac{2,043}{14,518}\\ =\frac{11,247}{14,518}\\ =0.77469\\ =0.7747\end{array}$

Thus, the probability that selected report is closed or comes from Queens is 0.7747.

c.

To determine

Compute the probability that selected report comes from Manhattan.

Answer to Problem 44E

The probability that selected report comes from Manhattan is 0.2858.

Explanation of Solution

Calculation:

The probability of the event E can be obtained by the formula:

$P\left(E\right)=\frac{\text{Number of outcomes in }E}{\text{Number of outcomes in}\text{sample space}}$

Here, event E denotes that selected report comes from Manhattan. Among the 14,518 reports; 4,149 of them are from Manhattan.

Substitute 4,149 for “number of outcomes in E” and 14,518 for “Number of outcomes in sample space” in the probability formula.

Therefore,

$\begin{array}{c}P\left(E\right)=\frac{4,149}{14,518}\\ =0.28578\\ =0.2858\end{array}$

Thus, the probability that selected report comes from Manhattan is 0.2858.

d.

To determine

Compute the probability that selected report does not comes from Manhattan.

Answer to Problem 44E

The probability that selected report does not comes from Manhattan is 0.7142.

Explanation of Solution

Calculation:

Complement Rule:

For any events E, $P\left(E^c\right)=1-P\left(E\right)$.

Event E denotes that selected report comes from Manhattan. Here events ‘selected report comes from Manhattan’ and ‘selected report does not comes from Manhattan’ are complements to each other.

From part (c) it is clear that, $P\left(E\right)=0.2858$. Substitute this values in the rule of complements.

Therefore,

$\begin{array}{c}P\left(\text{not comes from Manhattan}\right)=P\left(E^c\right)\\ =1-0.2858\\ =0.7142\end{array}$

Thus, the probability that selected report does not comes from Manhattan is 0.7142.

e.

To determine

Compute the probability that selected report is pending.

Answer to Problem 44E

The probability that selected report is pending is 0.0123.

Explanation of Solution

Calculation:

The probability of the event F can be obtained by the formula:

$P\left(F\right)=\frac{\text{Number of outcomes in }F}{\text{Number of outcomes in}\text{sample space}}$

Here, event F denotes that selected report is pending. Among the 14,518 reports; 178 of them are pending.

Substitute 178 for “number of outcomes in F” and 14,518 for “Number of outcomes in sample space” in the probability formula.

Therefore,

$\begin{array}{c}P\left(F\right)=\frac{178}{14,518}\\ =0.01226\\ =0.0123\end{array}$

Thus, the probability that selected report is pending is 0.0123.

f.

To determine

Compute the probability that selected report is from Bronx or Staten Island.

Answer to Problem 44E

The probability that selected report is from Bronx or Staten Island is 0.2083.

Explanation of Solution

Calculation:

Event  C denotes that selected report is from Bronx, event D denote that selected report is from Staten Island. Event C and D are mutually exclusive since the report comes from any one of the Borough.

The probability that a randomly selected report is from Bronx or Staten Island can be expressed as, $P\left(C\text{or }D\right)$.

Addition rule for mutually exclusive events:

For two mutually exclusive events C and D the addition rule states that

$P\left(C\text{or }D\right)=P\left(C\right)+P\left(D\right)$

From table, $P\left(C\right)=\frac{2,823}{14,518}\text{, and }P\left(D\right)=\frac{201}{14,518}$.

Substitute these values in the addition rule.

Therefore,

$\begin{array}{c}P\left(C\text{or }D\right)=\frac{2,823}{14,518}+\frac{201}{14,518}\\ =\frac{3,024}{14,518}\\ =0.20829\\ =0.2083\end{array}$

Thus, the probability that selected report is from Bronx or Staten Island is 0.2083.