Chapter 5.3, Problem 71E

a.

To determine

To find: the probability that the student is studying a language other than English.

Answer to Problem 71E

  $P\left(\text{studying another language}\right)=0.41$

Explanation of Solution

Given:

The table shows the percentage of student who study foreign language:

Language	Spanish	French	German	All others	None
Probability	0.26	0.09	0.03	0.03	0.59

Concept used:

Complement rule: $P\left(\text{not }A\right)=\left(1-P\left(A\right)\right)$ .

Calculation:

Using the complement rule to find the probability:

  $\begin{array}{l}P\left(\text{studying another language}\right)=1-P\left(\text{none}\right)\\ =1-0.59=0.41\end{array}$

Hence,the probability that the student is studying a language other than English is 0.41.

b.

To determine

To find: the conditional probability that a student is studying Spanish language.

Answer to Problem 71E

  $P\left(\text{Spanish}|\text{another language}\right)=0.6341$

Explanation of Solution

Given:

The student is studying Spanish language and is study some another language other than English. The table shows the percentage of student who study foreign-language:

Language	Spanish	French	German	All others	None
Probability	0.26	0.09	0.03	0.03	0.59

Concept used:

Probability is the number of favourable outcomes divided by the number of possible outcomes.

Thus, Definition condition probability:

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

Calculation:

Result from part (a): $P\left(\text{studying another language}\right)=0.41$

  $P\left(\text{another language and Spanish}\right)=P\left(\text{Spanish}\right)=0.26$

Thus, the required conditional probability becomes:

  $\begin{array}{l}P\left(\text{Spanish }/\text{another language}\right)=P\left(\text{Spanish and another language}\right)\\ =\frac{0.26}{0.41}\approx 0.6341\end{array}$

Hence,the conditional probability that a student is studying Spanish language is $P\left(\text{Spanish}|\text{another language}\right)=0.6341$