91影视

$2\%$bought midgrade gas.

$10\%$bought premium gas.

Customers paid with Credit card:

$28\%$ bought regular gasoline.

$34\%$bought midgrade gas.

$42\%$ bought premium gas.

The first level is:

There are three types of gasoline in the first level:

Regular, midgrade, and premium are the three levels of quality.

As a result, the first level necessitates three children:

Regular, midgrade, and premium are the three levels of quality.

Second tier:

There are two sorts of payment mechanisms at this level:

Using a credit card or not using a credit card (other payment method)

As a result, the second level has two children each child in the first level, i.e., Credit Card and No Credit Card.

The necessary tree diagram can be drawn as follows:

Fuel purchased by automobile drivers:

$88\%$ bought regular gasoline.

$2\%$ bought midgrade gas.

$10\%$ bought premium gas.

Customers paid with Credit card:

$28\%$ bought regular gasoline.

$34\%$ bought midgrade gas.

$42\%$ bought premium gas.

Let

R: Regular gasoline

M: Midgrade gas

P: Premium gas

C: Credit card

N: No Credit card

Now,

The corresponding probabilities:

Probability for the customer purchased regular gasoline,

$P\left(\mathrm{R}\right)=0.88$

Probability for the customer purchased midgrade gas,

$P\left(\mathrm{M}\right)=0.02$

Probability for the customer purchased premium gas,

$P\left(\mathrm{P}\right)=0.10$

Probability for the customer purchased regular gasoline paid with Credit card,

$P(C鈭�R)=0.28$

Probability for the customer purchased midgrade gas paid with Credit card,

$P(\mathrm{C}鈭�\mathrm{M})=0.34$

Probability for the customer purchased premium gas paid with Credit card,

$P(\mathrm{C}鈭�\mathrm{P})=0.42$

Apply general multiplication rule:

Probability for the customer paid with Credit card and purchased regular gasoline,

$P\left(\mathrm{C}\text{and}\mathrm{R}\right)=P\left(\mathrm{R}\right)脳P(\mathrm{C}鈭�\mathrm{R})=0.88脳0.28=0.2464$

Probability for the customer paid with Credit card and purchased midgrade gas,

$P\left(\mathrm{C}\text{and}\mathrm{M}\right)=P\left(\mathrm{M}\right)脳P(\mathrm{C}鈭�\mathrm{M})=0.02脳0.34=0.0068$

Probability for the customer paid with Credit card and purchased premium gas,

$P\left(\mathrm{C}\text{and}\mathrm{P}\right)=P\left(\mathrm{P}\right)脳P(\mathrm{C}鈭�\mathrm{P})=0.10脳0.42=0.0420$

Since the vehicles cannot be filled up with two types of gas at same time.

Apply the addition rule for mutually exclusive events:

$P\left(\mathrm{C}\right)=P\left(\mathrm{C}\text{and}\mathrm{R}\right)+P\left(\mathrm{C}\text{and}\mathrm{M}\right)+P\left(\mathrm{C}\text{and}\mathrm{P}\right)$

$=0.2464+0.0068+0.0420$

$=0.2952$

$Thus,ProbabilityforcustomerpaidwithaCreditcardis0.2952.$

Fuel purchased by automobile drivers:

$88\%$bought regular gasoline.

$2\%$ bought midgrade gas

$10\%$ bought premium gas

Customers paid with Credit card:

$28\%$bought regular gasoline.

$34\%$ bought midgrade gas

$42\%$ bought premium gas

From Part (b),

We have

Probability for the customer paid with a credit card,

$P\left(C\right)=0.2952$

Probability for the customer paid with credit card and purchased premium gas,

$P\left(\mathrm{C}\text{and}\mathrm{P}\right)=0.0420$

Apply the conditional probability:

$P(\mathrm{P}鈭�\mathrm{C})=\frac{P\left(\mathrm{C}\text{and}\mathrm{P}\right)}{P\left(\mathrm{C}\right)}=\frac{0.0420}{0.2952}=\frac{35}{246}鈮�0.1423$

Thus,

The probability for customer paid with credit card purchased premium gas is approx. $0.1423.$