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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, 9 = pizza available

The table shows the results of the simulation.
[tex]\[
\begin{tabular}{|l|l|l|l|l|l|}
\hline
Group number & 1 & 2 & 3 & 4 & 5 \\
\hline
Random digits & 19223 & 73676 & 45467 & 52711 & 95592 \\
\hline
\end{tabular}
\][/tex]
[tex]\[
\begin{tabular}{|l|l|l|l|l|l|}
\hline
Group number & 6 & 7 & 8 & 9 & 10 \\
\hline
Random digits & 68417 & 82739 & 60940 & 36009 & 38448 \\
\hline
\end{tabular}
\][/tex]

What is the estimated probability that Harriet will eat pizza for lunch every day next week?
A. [tex]$\frac{4}{10}=40 \%$[/tex]
B. [tex]$\frac{5}{10}=50 \%$[/tex]
C. [tex]$\frac{0}{10}=0 \%$[/tex]



Answer :

GinnyAnswer
To determine the estimated probability that Harriet will eat pizza for lunch every day next week, we need to examine the results of the simulation provided in the tables. Harriet's goal is to eat pizza every single day for a week, which consists of 5 days. The digits assigned in each group tell us whether pizza was available (digits 2 through 9) or not available (digits 0 and 1).

Here are the details of the groups and digits:

| Group Number | Random Digits |
|--------------|---------------|
| 1 | 19223 |
| 2 | 73676 |
| 3 | 45467 |
| 4 | 52711 |
| 5 | 95592 |
| 6 | 68417 |
| 7 | 82739 |
| 8 | 60940 |
| 9 | 36009 |
| 10 | 38448 |

To determine how many groups had pizza available every day, we need to check each group for the presence of digits '0' or '1'. If a group contains neither '0' nor '1', then pizza was available every day for that group.

Examining each group's digits:

1. Group 1 (19223): Contains '1', so no pizza every day.
2. Group 2 (73676): Contains no '0' or '1', so pizza every day.
3. Group 3 (45467): Contains no '0' or '1', so pizza every day.
4. Group 4 (52711): Contains '1', so no pizza every day.
5. Group 5 (95592): Contains no '0' or '1', so pizza every day.
6. Group 6 (68417): Contains '1', so no pizza every day.
7. Group 7 (82739): Contains no '0' or '1', so pizza every day.
8. Group 8 (60940): Contains '0', so no pizza every day.
9. Group 9 (36009): Contains '0', no pizza every day.
10. Group 10 (38448): Contains no '0' or '1', so pizza every day.

Based on the above examination, the groups where Harriet had pizza every day are Groups 2, 3, 5, 7, and 10. That makes a total of 5 groups.

The total number of groups is 10. Therefore, the estimated probability that Harriet will eat pizza for lunch every day next week is calculated as follows:

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

Converting this fraction to percentage:

[tex]\[ 0.5 \times 100\% = 50\% \][/tex]

Thus, the correct answer is:
B. [tex]\(\frac{5}{10} = 50\%\)[/tex]

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