Chapter 5, Problem T5.13SPT

(a)

To determine

To find the probability that the customer will pay less than the usual cost of the buffet.

Answer to Problem T5.13SPT

  $63.98\%$ .

Explanation of Solution

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 $23\text{ out of 36}$ outcomes are below $12.99$ and thus we pay less than $\$12.99\text{ in 23 of the 36}$ outcomes. Thus, we have the probability of that the customer will pay less than the usual cost of the buffet as:

  $\begin{array}{l}P(\text{Pay less than the usual cost})=\frac{23}{36}\\ =0.6389\\ =63.98\%\end{array}$

(b)

To determine

To find the probability that all $4$ of these friends end up paying less than the usual cost of the buffet.

Answer to Problem T5.13SPT

  $16.66\%$ .

Explanation of Solution

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 $23\text{ out of 36}$ outcomes are below $12.99$ and thus we pay less than $\$12.99\text{ in 23 of the 36}$ outcomes. Thus, we have the probability of that the customer will pay less than the usual cost of the buffet as:

  $\begin{array}{l}P(\text{Pay less than the usual cost})=\frac{23}{36}\\ =0.6389\\ =63.98\%\end{array}$

Let A be one person pays less than the usual cost and B be $4\text{ out of 4}$ people pay less than the usual cost.

So, $P(A)=0.6389$ .

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{array}{l}P(B)=P(A)\times P(A)\times P(A)\times P(A)\\ $ ={(P(A))}^4$={0.6389}^4\\ =0.1666\\ =16.66\%\end{array}$

(c)

To determine

To find the probability that at least $1\text{ of the 4}$ friends end up paying more than the usual cost of the buffet.

Answer to Problem T5.13SPT

  $83.34\%$ .

Explanation of Solution

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 $23\text{ out of 36}$ outcomes are below $12.99$ and thus we pay less than $\$12.99\text{ in 23 of the 36}$ outcomes. Thus, we have the probability of that the customer will pay less than the usual cost of the buffet as:

  $\begin{array}{l}P(\text{Pay less than the usual cost})=\frac{23}{36}\\ =0.6389\\ =63.98\%\end{array}$

Let A be one person pays less than the usual cost and $B^c$ be at least $1\text{ of the 4}$ friends end up paying more than the usual cost of the buffet.

So, $P(A)=0.6389$ .

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{array}{l}P(B^c)=P(A)\times P(A)\times P(A)\times P(A)\\ $ ={(P(A))}^4$={0.6389}^4\\ =0.1666\end{array}$

Now, by using the complement rule we have,

  $\begin{array}{l}P(B)=P({(}^{B^c})\\ =1-P(B^c)\\ =1-0.1666\\ =0.8334\\ =83.34\%\end{array}$