Answer :
We are given the following partial table with information about students’ pets and weekly screen time:
[tex]$$
\begin{array}{|c|c|c|c|}
\hline \text{Screen time, weekly} & \text{Pets} & \text{No pets} & \text{Total} \\
\hline \text{At most 20 hours} & 15 & \,\,?\,\, & \,\,?\,\, \\
\hline \text{Over 20 hours} & \,\,?\,\, & \,\,?\,\, & 13 \\
\hline \text{Total} & 21 & 24 & \,\,?\,\, \\
\hline
\end{array}
$$[/tex]
We will complete the table step by step and then compute the required conditional probabilities.
–––––––––––
Step 1. Completing the Table
1.1. Find the number of students with pets and over 20 hours of screen time:
Since the total number of students with pets is 21 and 15 of them have at most 20 hours of screen time, the number with pets and over 20 hours is
[tex]$$
21 - 15 = 6.
$$[/tex]
1.2. Find the number of students with no pets and over 20 hours of screen time:
The total number of students with over 20 hours is given as 13. Out of these, 6 have pets, so the remaining students with no pets are
[tex]$$
13 - 6 = 7.
$$[/tex]
1.3. Find the number of students with no pets and at most 20 hours of screen time:
The total number of students with no pets is 24. Since 7 of them have over 20 hours, the number with at most 20 hours is
[tex]$$
24 - 7 = 17.
$$[/tex]
1.4. Find the row total for “At most 20 hours”:
Now, add the number of students with pets and no pets for this screen time category:
[tex]$$
15 + 17 = 32.
$$[/tex]
1.5. Finally, find the overall total number of students:
Add the totals of the two screen time rows:
[tex]$$
32 + 13 = 45.
$$[/tex]
The completed table is:
[tex]$$
\begin{array}{|c|c|c|c|}
\hline \text{Screen time, weekly} & \text{Pets} & \text{No pets} & \text{Total} \\
\hline \text{At most 20 hours} & 15 & 17 & 32 \\
\hline \text{Over 20 hours} & 6 & 7 & 13 \\
\hline \text{Total} & 21 & 24 & 45 \\
\hline
\end{array}
$$[/tex]
–––––––––––
Step 2. Calculating the Required Probabilities
2.1. Probability that a student who does not have pets had over 20 hours of weekly screen time:
For students with no pets, there are 24 in total. Out of these, 7 have over 20 hours of screen time. The conditional probability is given by
[tex]$$
P(\text{Over 20 hours} \mid \text{No pets}) = \frac{7}{24} \approx 0.29167.
$$[/tex]
2.2. Probability that a student who had over 20 hours of weekly screen time has pets:
For the group of students with over 20 hours of screen time, there are 13 in total. Among these, 6 have pets. This gives the probability
[tex]$$
P(\text{Pets} \mid \text{Over 20 hours}) = \frac{6}{13} \approx 0.46154.
$$[/tex]
–––––––––––
Final Answers:
- The probability that a student who does not have pets had over 20 hours of weekly screen time is [tex]$$\frac{7}{24} \approx 0.29167.$$[/tex]
- The probability that a student who had over 20 hours of weekly screen time has pets is [tex]$$\frac{6}{13} \approx 0.46154.$$[/tex]
[tex]$$
\begin{array}{|c|c|c|c|}
\hline \text{Screen time, weekly} & \text{Pets} & \text{No pets} & \text{Total} \\
\hline \text{At most 20 hours} & 15 & \,\,?\,\, & \,\,?\,\, \\
\hline \text{Over 20 hours} & \,\,?\,\, & \,\,?\,\, & 13 \\
\hline \text{Total} & 21 & 24 & \,\,?\,\, \\
\hline
\end{array}
$$[/tex]
We will complete the table step by step and then compute the required conditional probabilities.
–––––––––––
Step 1. Completing the Table
1.1. Find the number of students with pets and over 20 hours of screen time:
Since the total number of students with pets is 21 and 15 of them have at most 20 hours of screen time, the number with pets and over 20 hours is
[tex]$$
21 - 15 = 6.
$$[/tex]
1.2. Find the number of students with no pets and over 20 hours of screen time:
The total number of students with over 20 hours is given as 13. Out of these, 6 have pets, so the remaining students with no pets are
[tex]$$
13 - 6 = 7.
$$[/tex]
1.3. Find the number of students with no pets and at most 20 hours of screen time:
The total number of students with no pets is 24. Since 7 of them have over 20 hours, the number with at most 20 hours is
[tex]$$
24 - 7 = 17.
$$[/tex]
1.4. Find the row total for “At most 20 hours”:
Now, add the number of students with pets and no pets for this screen time category:
[tex]$$
15 + 17 = 32.
$$[/tex]
1.5. Finally, find the overall total number of students:
Add the totals of the two screen time rows:
[tex]$$
32 + 13 = 45.
$$[/tex]
The completed table is:
[tex]$$
\begin{array}{|c|c|c|c|}
\hline \text{Screen time, weekly} & \text{Pets} & \text{No pets} & \text{Total} \\
\hline \text{At most 20 hours} & 15 & 17 & 32 \\
\hline \text{Over 20 hours} & 6 & 7 & 13 \\
\hline \text{Total} & 21 & 24 & 45 \\
\hline
\end{array}
$$[/tex]
–––––––––––
Step 2. Calculating the Required Probabilities
2.1. Probability that a student who does not have pets had over 20 hours of weekly screen time:
For students with no pets, there are 24 in total. Out of these, 7 have over 20 hours of screen time. The conditional probability is given by
[tex]$$
P(\text{Over 20 hours} \mid \text{No pets}) = \frac{7}{24} \approx 0.29167.
$$[/tex]
2.2. Probability that a student who had over 20 hours of weekly screen time has pets:
For the group of students with over 20 hours of screen time, there are 13 in total. Among these, 6 have pets. This gives the probability
[tex]$$
P(\text{Pets} \mid \text{Over 20 hours}) = \frac{6}{13} \approx 0.46154.
$$[/tex]
–––––––––––
Final Answers:
- The probability that a student who does not have pets had over 20 hours of weekly screen time is [tex]$$\frac{7}{24} \approx 0.29167.$$[/tex]
- The probability that a student who had over 20 hours of weekly screen time has pets is [tex]$$\frac{6}{13} \approx 0.46154.$$[/tex]