At Dicey Dave’s Diner, the dinner buffet usually costs $12.99. Once a month, Dave sponsors “lucky buffet” night. On that night, each patron can either pay the usual price or roll two fair, six-sided dice and pay a number of dollars equal to the product of the numbers showing on the two faces. The table shows the sample space of this chance process.

a. A customer decides to play Dave’s “lucky buffet” game. Find the probability that the customer will pay less than the usual cost of the buffet.

b. A group of 4 friends comes to Dicey Dave’s Diner to play the “lucky buffet” game. Find the probability that all 4 of these friends end up paying less than the usual cost of the buffet.

c. Find the probability that at least 1 of the 4 friends ends up paying more than the usual cost of the buffet

Dave sponsors a "lucky buffet" night once a month, according to the question. On that night, each patron has the option of paying the regular price or rolling two fair dice and paying the sum of the numbers on the two dice. The sample space of the chance process is shown in the table in the question.

In the question it is given that once a month, Dave sponsors “lucky buffet” night. On that night each patron can either pay the usual price or roll two fair dice and pay a number of dollars equal to the product of the numbers showing on the two dice. The table in the question is given which shows the sample space of the chance process. Thus, the table contains possible outcomes. We also note that $23outof36$outcomes are below $12.99$and thus we pay less than $\$12.99in23ofthe36$outcomes. Thus, we have the probability of that the customer will pay less than the usual cost of the buffet as:

$P(Paylessthantheusual\cos t)=\frac{23}{36}$

$$\begin{gathered} =0.6389 \\ =63.98\% \end{gathered}$$

In the question it is given that once a month, Dave sponsors “lucky buffet” night. On that night each patron can either pay the usual price or roll two fair dice and pay a number of dollars equal to the product of the numbers showing on the two dice. The table in the question is given which shows the sample space of the chance process. Thus, the table contains possible outcomes. We also note that$23outof36$outcomes are below$12.99$and thus we pay less than$\$12.99in23ofthe36$outcomes. Thus, we have the probability of that the customer will pay less than the usual cost of the buffet as:

$$\begin{gathered} P(Paylessthantheusual\cos t)=\frac{23}{36} \\ =0.6389 \\ =63.98\% \end{gathered}$$

Let A be one person pays less than the usual cost and B be $4outof4$people pay less than the usual cost.

So, $P\left(A\right)=0.6389$

Since the rolls of the dice for the different people are independent of each other we can use the multiplication rule for independent events as:

$$\begin{gathered} P\left(B\right)=P\left(A\right)\times P\left(A\right)\times P\left(A\right)\times P\left(A\right) \\ =(P{\left(A\right))}^4=0.{6389}^4 \\ =0.1666 \\ =16.66\% \end{gathered}$$

In the question it is given that once a month, Dave sponsors “lucky buffet” night. On that night each patron can either pay the usual price or roll two fair dice and pay a number of dollars equal to the product of the numbers showing on the two dice. The table in the question is given which shows the sample space of the chance process. Thus, the table contains $36$possible outcomes. We also note that $23outof36$outcomes are below $12.99$and thus we pay less than $\$12.99in23ofthe36$outcomes. Thus, we have the probability of that the customer will pay less than the usual cost of the buffet as:

$$\begin{gathered} P(Paylessthantheusual\cos t)=\frac{23}{36} \\ =0.6389 \\ =63.98\% \end{gathered}$$

Let A be one person pays less than the usual cost and $B^c$be at least $1ofthe4$friends end up paying more than the usual cost of the buffet.

So, $P\left(A\right)=0.6389$

Since the rolls of the dice for the different people are independent of each other we can use the multiplication rule for independent events as:

$$\begin{gathered} P\left(B^c\right)=P\left(A\right)\times P\left(A\right)\times P\left(A\right)\times P\left(A\right) \\ =(P{\left(A\right))}^4 \\ =0.{6389}^4 \\ =0.1666 \end{gathered}$$

Now, by using the complement rule we have,

$$\begin{gathered} P\left(B\right)=P\left({\left(B^c\right)}^c\right)=1-P\left(B^c\right) \\ =1-0.1666 \\ =0.8334 \\ =83.34\% \end{gathered}$$