Chapter 0.6, Problem 15E

(a)

To determine

Probability for Paige got passed took the class.

Answer to Problem 15E

Probability that Paige got passed took the class is approx. 0.78.

Explanation of Solution

Given information:

Students’ performance in their first driving test:

Status	Class	No Class
passed	64	48
failed	18	32

Calculation:

According to the conditional probability,

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

Calculate the table total:

Status	Class	No Class	Total
passed	64	48	112
failed	18	32	50
Total	82	80	162

Note that

The information about 162 students is provided in the table.

Thus,

The number of possible outcomes is 162.

Also note that

In the table, 64 of the 162 students got passed took the class.

Thus,

The number of favorable outcomes is 64.

When the number of favorable outcomes is divided by the number of possible outcomes, we get the probability.

  $P\left(PassandClass\right)=\frac{\text{Number}\text{of}\text{favourable}\text{outcomes}}{\text{Number}\text{of}\text{possible}\text{outcomes}}=\frac{64}{162}$

Now,

Note that

In the table, 82 of 162 students took class.

In this case, the number of favorable outcomes is 82 and number of possible outcomes is 162.

  $P\left(Class\right)=\frac{\text{Number}\text{of}\text{favourable}\text{outcomes}}{\text{Number}\text{of}\text{possible}\text{outcomes}}=\frac{82}{162}$

Apply the conditional probability:

  $\begin{array}{l}P\left(Pass|Class\right)=\frac{P\left(PassandClass\right)}{P\left(Class\right)}\\ =\frac{\frac{64}{162}}{\frac{82}{162}}\\ =\frac{64}{82}\\ =\frac{32}{41}\\ \approx 0.78\end{array}$

Thus,

The conditional probability for Paige got passed took the class is approx. 0.78.

(b)

To determine

Probability for Madison got failed did not take the class.

Answer to Problem 15E

Probability that Madison got failed did not take the class is 0.4.

Explanation of Solution

Given information:

Students’ performance in their first driving test:

Status	Class	No Class
passed	64	48
failed	18	32

Calculation:

According to the conditional probability,

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

Calculate the table total:

Status	Class	No Class	Total
passed	64	48	112
failed	18	32	50
Total	82	80	162

Note that

The information about 162 students is provided in the table.

Thus,

The number of possible outcomes is 162.

Also note that

In the table, 32 of the 162 students got failed didn’t take the class.

Thus,

The number of favorable outcomes is 32.

When the number of favorable outcomes is divided by the number of possible outcomes, we get the probability.

  $P\left(FailandNoClass\right)=\frac{\text{Number}\text{of}\text{favourable}\text{outcomes}}{\text{Number}\text{of}\text{possible}\text{outcomes}}=\frac{32}{162}$

Now,

Note that

In the table, 80 of 162 students took no class.

In this case, the number of favorable outcomes is 80 and number of possible outcomes is 162.

  $P\left(NoClass\right)=\frac{\text{Number}\text{of}\text{favourable}\text{outcomes}}{\text{Number}\text{of}\text{possible}\text{outcomes}}=\frac{80}{162}$

Apply the conditional probability:

  $\begin{array}{l}P\left(Fail|NoClass\right)=\frac{P\left(FailandNoClass\right)}{P\left(NoClass\right)}\\ =\frac{\frac{32}{162}}{\frac{80}{162}}\\ =\frac{32}{80}\\ =\frac{4}{10}\\ =0.4\end{array}$

Thus,

The conditional probability for Madison got failed didn’t took the class is 0.4.

(c)

To determine

Probability for Jamal did not take the class got passed.

Answer to Problem 15E

Probability that Jamal did not take the class got passed is approx. 0.43.

Explanation of Solution

Given information:

Students’ performance in their first driving test:

Status	Class	No Class
passed	64	48
failed	18	32

Calculation:

According to the conditional probability,

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

Calculate the table total:

Status	Class	No Class	Total
passed	64	48	112
failed	18	32	50
Total	82	80	162

Note that

The information about 162 students is provided in the table.

Thus,

The number of possible outcomes is 162.

Also note that

In the table, 48 of the 162 students got passed didn’t take the class.

Thus,

The number of favorable outcomes is 48.

When the number of favorable outcomes is divided by the number of possible outcomes, we get the probability.

  $P\left(PassandNoClass\right)=\frac{\text{Number}\text{of}\text{favourable}\text{outcomes}}{\text{Number}\text{of}\text{possible}\text{outcomes}}=\frac{48}{162}$

Now,

Note that

In the table, 112 of 162 students got passed.

In this case, the number of favorable outcomes is 112 and number of possible outcomes is 162.

  $P\left(Pass\right)=\frac{\text{Number}\text{of}\text{favourable}\text{outcomes}}{\text{Number}\text{of}\text{possible}\text{outcomes}}=\frac{112}{162}$

Apply the conditional probability:

  $\begin{array}{l}P\left(NoClass|Pass\right)=\frac{P\left(NoClassandPass\right)}{P\left(Pass\right)}\\ =\frac{\frac{48}{162}}{\frac{112}{162}}\\ =\frac{48}{112}\\ =\frac{3}{7}\\ \approx 0.43\end{array}$

Thus,

The conditional probability for Jamal took no class got passed is approx. 0.43.