Chapter 5, Problem 10RE

Heart attack: The following table presents the number of hospitalizations for myocardial infarction (heart attack) for men and women in various age groups.

1. What is the probability that a randomly chosen patient is a woman?
2. What is the probability that a randomly chosen patient is aged 45-64?
3. What is the probability that a randomly chosen patient is a woman and aged 45-64?
4. What is the probability that a randomly chosen patient is a woman or aged 45-64?
5. What is the probability that a randomly chosen patient is a woman given that the patient is aged 45-64?
6. What is the probability that a randomly chosen patient is aged 45-64 given that the patient is a woman?

To determine

To find: The probability that a randomly chosen patient is a woman.

Answer to Problem 10RE

  $0.4035$

Explanation of Solution

Given information:

  $\boxed{\begin{array}{llll}Age\hfill & Male\hfill & Female\hfill & Total\hfill \\ 18-44\hfill & 26828\hfill & 9265\hfill & 36093\hfill \\ 45-64\hfill & 166340\hfill & 68666\hfill & 235006\hfill \\ 65-84\hfill & 155707\hfill & 124289\hfill & 279996\hfill \\ 85\text{ and above}\hfill & 35524\hfill & 57785\hfill & 93309\hfill \\ Total\hfill & 384399\hfill & 260005\hfill & 644404\hfill \end{array}}$

Calculation:

Out of $644404$ hospitalizations, there were $260005$ women.

Hence,

  $\begin{array}{l}P(\text{Woman})=\frac{260005}{644404}\\ =0.4035\end{array}$

To determine

To find: The probability that a randomly chosen patient is aged $45-64$ .

Answer to Problem 10RE

  $0.3647$

Explanation of Solution

Calculation:

Out of $644404$ hospitalizations, there were $235006$ aged $45-64$ .

Hence,

  $\begin{array}{l}P(\text{aged }45-64)=\frac{235006}{644404}\\ =0.3647\end{array}$

To determine

To find: The probability that a randomly chosen patient is a woman and aged $45-64$ .

Answer to Problem 10RE

  $0.1066$

Explanation of Solution

Calculation:

Out of $644404$ patients, there were $68666$ woman andaged $45-64$ .

Therefore,

  $\begin{array}{l}P(\text{woman and aged }45-64)=\frac{68666}{644404}\\ =0.1066\end{array}$

To determine

To find: The probability that a randomly chosen patient is a woman or aged $45-64$ .

Answer to Problem 10RE

  $0.6616$

Explanation of Solution

The general addition rule for any two events $A$ and $B$ is defined as,

  $P(A\text{or}B)=P(A)+P(B)-P(A\text{and }B)$

Here,

  $\begin{array}{l}P(\text{Woman}\text{or}\text{aged }45-64)=P(\text{Woman})+P(\text{aged }45-64)-P\left(\begin{array}{c}\text{Woman}\text{and}\\ \text{aged }45-64\end{array}\right)\\ =\frac{260005}{644404}+\frac{235006}{644404}-\frac{68666}{644404}\\ =\frac{426345}{644404}=0.6616\end{array}$

To determine

To find: The probability that a randomly chosen patient is a women given that the patient is aged $45-64$ .

Answer to Problem 10RE

  $0.2922$

Explanation of Solution

Calculation:

The probability of $B$ given $A$ is defined as,

  $P(B|A)=\frac{P(A\text{ and }B)}{P(A)}$

Let,

  $P(B|A)=P(\text{women given that the patient is aged }45-64)$

So

  $\begin{array}{l}P(B\text{ and }A)=P(\text{ aged }45-64\text{ and}\text{woman})=\frac{68666}{644404}\\ P(A)=P(\text{aged }45-64)=\frac{235006}{644404}\end{array}$

Hence

  $P(B|A)=P(\text{women given that the patient is aged }45-64)=\frac{68666}{235006}=0.2922$

To determine

To find: The probability that a randomly chosen patient is aged $45-64$ given that the patient is woman.

Answer to Problem 10RE

  $0.2641$

Explanation of Solution

Calculation:

The probability of $A$ given $B$ is defined as,

  $P(A|B)=\frac{P(B\text{ and }A)}{P(B)}$

Let,

  $P(A|B)=P(\text{aged }45-64\text{ given that the patient is a women})$

So

  $\begin{array}{l}P(A\text{ and }B)=P(\text{ woman and aged }45-64)=\frac{68666}{644404}\\ P(B)=P(\text{Woman})=\frac{260005}{644404}\end{array}$

Hence

  $P(A|B)=P(\text{aged }45-64\text{ given that the patient is a women})=\frac{68666}{235006}=0.2641$