Chapter 15, Problem 5E

(a)

To determine

To find: the probability that the person is from the United States.

Answer to Problem 5E

0.2025

Explanation of Solution

Given:

  

Formula used:

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

Calculation:

  $P\left(USA\right)=\frac{1557}{7690}=0.2025$

(b)

To determine

To find: the probability that the person completed his or her education before college.

Answer to Problem 5E

0.696

Explanation of Solution

Given:

  

Formula used:

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

Calculation:

P (some HS or primary and less) = P(some HS) + P(primary and less) − P(some HS and primary and less) = $\frac{4195}{7690}+\frac{1161}{7690}+0=0.696$

(c)

To determine

To find: the probability that the person is from France or did some post-graduate study.

Answer to Problem 5E

0.2404

Explanation of Solution

Given:

  

Formula used:

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

Calculation:

P (France or post-grad) = P(France) + P(Post-grad) − P(France and post-grad) = $\frac{1557}{7690}+\frac{379}{7690}-\frac{69}{7690}=\frac{1849}{7690}=0.2404$

(d)

To determine

To find: the probability that the person is from France and finished only primary school or less.

Answer to Problem 5E

0.0402

Explanation of Solution

Given:

  

Formula used:

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

Calculation:

P (France and primary or less) = $\frac{309}{7690}=0.0402$