A researcher randomly sampled 222 high school students to determine their favorite color and whether or not they played a sport. The two-way table displays the data.

Favorite Color
\begin{tabular}{|c|c|c|c|c|c|}
\cline {2-6}
\multicolumn{1}{c|}{} & \begin{tabular}{c}
Red, \\ Yellow, or \\ Orange
\end{tabular} & \begin{tabular}{c}
Blue, \\ Green, or \\ Purple
\end{tabular} & \begin{tabular}{c}
Pink or \\ Violet
\end{tabular} & Other & Total \\
\hline
Plays Sports & 32 & 36 & 25 & 11 & 104 \\
\hline
Does Not Play Sports & 35 & 41 & 13 & 29 & 118 \\
\hline
Total & 67 & 77 & 38 & 40 & 222 \\
\hline
\end{tabular}

A randomly selected student who participated in the survey is selected. Let event [tex]$S=$[/tex] the student plays a sport and let event [tex]$B =$[/tex] favorite color is blue, green, or purple. What is the value of [tex]$P(B \mid S )$[/tex]?



Answer :

To determine the value of [tex]\( P(B \mid S) \)[/tex], which is the probability that a randomly selected student who plays a sport has blue, green, or purple as their favorite color, we can use the following approach:

1. Identify Total Number of Students Who Play Sports:
- From the table, we see that the total number of students who play sports is given in the "Plays Sports" row. This total is 104 students.

2. Identify Number of Students Who Play Sports and Prefer Blue, Green, or Purple:
- From the same row, the number of students who play sports and prefer blue, green, or purple is 36 students.

3. Calculate the Conditional Probability:
- Conditional probability [tex]\( P(B \mid S) \)[/tex] is calculated as the number of students who play sports and prefer blue, green, or purple ([tex]\( 36 \)[/tex]) divided by the total number of students who play sports ([tex]\( 104 \)[/tex]).

The calculation should then look like this:

[tex]\[ P(B \mid S) = \frac{\text{Number of students who play sports and prefer blue, green, or purple}}{\text{Total number of students who play sports}} = \frac{36}{104} \][/tex]

4. Simplify the fraction:
- [tex]\( P(B \mid S) = \frac{36}{104} \approx 0.34615384615384615 \)[/tex]

Therefore, the probability [tex]\( P(B \mid S) \)[/tex] is approximately [tex]\( 0.346 \)[/tex] or [tex]\( 34.6\% \)[/tex].