Fill ’err up! In a recent month, $88\%$of automobile drivers filled their vehicles with regular gasoline, $2\%$purchased midgrade gas, and $10\%$bought premium gas.$17$ Of those who bought regular gas, $28\%$ paid with a credit card; of customers who bought midgrade and premium gas, $34\%$ and $42\%$, respectively, paid with a credit card. Suppose we select a customer at random.

Draw a tree diagram to represent this situation. What’s the probability that the customer paid with a credit card? Use the four-step process to guide your work.

Regular gasoline was utilized by $88$percent of the vehicles, with midgrade gas purchased by$2\%$ and premium gas purchased by $10\%$ Regular gas is bought with a credit card $28\%$ of the time, midgrade gas is bought with a credit card $34\%$ of the time, and premium gas is bought with a credit card $42\%$of the time.

Conditional probability: $P(A\cap B)=P(A/B)\times P\left(B\right)$

The tree diagram is shown below, according to the question:

Now, for each type of gasoline, determine the likelihood of paying using a credit card.

$P(Regular\cap Creditcard)=P(Creditcard/Regular)\times P\left(Regular\right)=88\%\times 28\%=0.88\times 0.28=0.2464$

$$\begin{gathered} P(Midgrade\cap Creditcard)=P(Creditcard/Midgrade)\times P\left(Midgrade\right) \\ =2\%\times 34\% \\ =0.02\times 0.34=0.0068 \end{gathered}$$

$$\begin{gathered} P(Premium\cap Creditcard)=P(Creditcard/Premium)\times P\left(Premium\right) \\ =10\%\times 42\% \\ =0.10\times 0.42 \\ =0.042 \end{gathered}$$

Fill in the blanks with the corresponding probabilities:

As a result, the likelihood of the buyer paying using a credit card is $0.2952\%$