Answered

Harriet buys lunch at school every day. She always gets pizza when it is available. The cafeteria has pizza about [tex]80\%[/tex] of the time.

Harriet runs a simulation to model this using a random number generator. She assigns these digits to the possible outcomes for each day of the week:
- Let 0 and 1 = no pizza available
- Let 2, 3, 4, 5, 6, 7, 8, and 9 = pizza available

The table shows the results of the simulation.

\begin{tabular}{|l|l|l|l|l|l|}
\hline Group number & 1 & 2 & 3 & 4 & 5 \\
\hline Random digits & 19223 & 73676 & 45467 & 52711 & 95592 \\
\hline
\end{tabular}

\begin{tabular}{|l|l|l|l|l|l|}
\hline Group number & 6 & 7 & 8 & 9 & 10 \\
\hline Random digits & 68417 & 82739 & 60940 & 36009 & 38448 \\
\hline
\end{tabular}

What is the estimated probability that Harriet will eat pizza for lunch every day next week?



Answer :

GinnyAnswer
To estimate the probability that Harriet will eat pizza for lunch every day next week, we can model the situation using the random digit simulations provided. Let's break down the solution step-by-step:

1. Group Setup:
Harriet's simulation consists of 10 groups of 5-digit numbers, each representing a week of school days. The groups and their corresponding random digits are:

[tex]\[ \begin{aligned} &1: 19223, &2: 73676, &3: 45467, &4: 52711, &5: 95592,\\ &6: 68417, &7: 82739, &8: 60940, &9: 36009, &10: 38448. \end{aligned} \][/tex]

2. Defining Pizza Availability:
Pizza is available on days represented by the digits 2, 3, 4, 5, 6, 7, 8, and 9. On days represented by the digits 0 or 1, pizza is not available.

3. Evaluating Each Group:
We need to evaluate each 5-digit number and check if pizza is available on all 5 days of the week to consider that week as successful (having pizza every day).

- Group 1 (19223): Pizza available digits are present in all positions.
- Group 2 (73676): Pizza available digits are present in all positions.
- Group 3 (45467): Pizza available digits are present in all positions.
- Group 4 (52711): The digit 1 (no pizza) is present; hence, it's not successful.
- Group 5 (95592): Pizza available digits are present in all positions.
- Group 6 (68417): The digit 1 (no pizza) is present; hence, it's not successful.
- Group 7 (82739): Pizza available digits are present in all positions.
- Group 8 (60940): Pizza available digits are present in all positions.
- Group 9 (36009): Pizza available digits are present in all positions.
- Group 10 (38448): Pizza available digits are present in all positions.

4. Counting Successful Weeks:
The successful groups are: 1, 2, 3, 5, 7, and 10. Thus, there are 6 successful simulations.

5. Total Simulations:
There are a total of 10 simulations (groups).

6. Calculating the Estimated Probability:
The estimated probability is the number of successful simulations divided by the total number of simulations.

[tex]\[ \text{Probability} = \frac{\text{Number of successful simulations}}{\text{Total number of simulations}} = \frac{5}{10} = 0.5. \][/tex]

Therefore, the estimated probability that Harriet will eat pizza for lunch every day next week is [tex]\(0.5\)[/tex] or [tex]\(50\%\)[/tex].

Other Questions