Chapter 15, Problem 6E

(a)

To determine

To find: the probability that the person is a Human Ecology Student.

Answer to Problem 6E

0.193

Explanation of Solution

Given:

  

Formula used:

  $\text{Probability}\text{=}\frac{\text{Favourable Cases}}{\text{Total}\text{Cases}}$

Calculation:

  $P\left(\text{human ecology student}\right)=\frac{43}{223}=0.193$

(b)

To determine

To find: the probability that the person is a firstborn student.

Answer to Problem 6E

0.507

Explanation of Solution

Given:

  

Formula used:

  $\text{Probability}\text{=}\frac{\text{Favourable Cases}}{\text{Total}\text{Cases}}$

Calculation:

  $P\left(\text{firstborn student}\right)=\frac{113}{223}=0.507$

(c)

To determine

To find: the probability that the person is firstborn and a Human Ecology student.

Answer to Problem 6E

0.067

Explanation of Solution

Given:

  

Formula used:

  $\text{Probability}\text{=}\frac{\text{Favourable Cases}}{\text{Total}\text{Cases}}$

Calculation:

  $P\left(\text{firstborn or human ecology}\right)=\frac{15}{223}=0.067$

(d)

To determine

To find: the probability that the person is firstborn or a Human Ecology student

Answer to Problem 6E

0.632

Explanation of Solution

Given:

  

Formula used:

  $\text{Probability}\text{=}\frac{\text{Favourable Cases}}{\text{Total}\text{Cases}}$

Calculation:

  $\begin{array}{c}P\left(\text{firstborn or human ecology}\right)=P\left(\text{first born}\right)+P\left(\text{human ecology}\right)-P\left(\text{first born and human ecology}\right)\\ =\frac{113}{223}+\frac{43}{223}-\frac{15}{223}\\ =0.632\end{array}$