Chapter 5.3, Problem 64E

a.

To determine

To find:the probability that the result for “he or she survived” in the given table.

Answer to Problem 64E

  $P\left(\text{surived}/\text{First class}\right)=0.6167$ .

Explanation of Solution

Given:

The two-way table gives information about adult passenger who lives and who died as in the given table:

Class of Travel	Survived	Died
First class	197	122
Second class	94	167
Third class	151	476

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:

According to the question, the person is selected first class. Now to find the probability that the result for “he or she survived”.

Therefore,

  $P\left(\text{First class and surive}\right)=\frac{\text{favourable outcomes}}{\text{possible outcomes}}=\frac{197}{1207}$

  $P\left(\text{First class}\right)=\frac{\text{favourable outcomes}}{\text{possible outcomes}}=\frac{197+122}{1207}=\frac{319}{1207}$

Thus, the conditional probability becomes:

  $\begin{array}{l}P\left(\text{surived}/\text{First class}\right)=\frac{P\left(\text{First class and survived}\right)}{P\left(\text{First class}\right)}\\ =\frac{197/1207}{319/1207}=\frac{197}{319}\approx 0.6167\end{array}$

Hence, the probability that the result for “he or she survived” is $P\left(\text{surived}/\text{First class}\right)=0.6167$ .

b.

To determine

To find: the probability that the result for “he or she survived” in the given table.

Answer to Problem 64E

  $P\left(\text{Third class/survived}\right)=0.3416$ .

Explanation of Solution

Given:

The two-way table gives information about adult passenger who lives and who died as in the given table:

Class of Travel	Survived	Died
First class	197	122
Second class	94	167
Third class	151	476

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:

According to the question, the person is selecteted for survival. Now to find the probability that the result for “he or she was a third-class passenger”.

Therefore,

  $P\left(\text{Third class and surive}\right)=\frac{\text{favourable outcomes}}{\text{possible outcomes}}=\frac{151}{1207}$

  $P\left(\text{First class}\right)=\frac{\text{favourable outcomes}}{\text{possible outcomes}}=\frac{197+94+151}{1207}=\frac{442}{1207}$

Thus, the conditional probability becomes:

  $\begin{array}{l}P\left(\text{Third class/survived}\right)=\frac{P\left(\text{Third class and survived}\right)}{P\left(\text{First class}\right)}\\ =\frac{151/1207}{442/1207}=\frac{151}{442}\approx 0.3416\end{array}$

Hence, the probability that the result for “he or she was a third-class passenger”is

  $P\left(\text{Third class/survived}\right)=0.3416$ .