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Answer :
We are asked to compute the conditional probability that a dormitory resident has had a flu shot given that the resident is male. Conditional probability is calculated using the formula
[tex]$$
P(A \mid B) = \frac{P(A \cap B)}{P(B)},
$$[/tex]
where in this context:
- [tex]$A$[/tex] is the event "resident had a flu shot",
- [tex]$B$[/tex] is the event "resident is male."
From the table, we have:
- The number of male residents who had a flu shot is [tex]$39$[/tex].
- The total number of male residents is [tex]$51$[/tex].
Thus, the probability that a male resident had a flu shot is
[tex]$$
P(\text{Flu Shot} \mid \text{Male}) = \frac{39}{51}.
$$[/tex]
We can simplify the fraction by dividing the numerator and the denominator by their greatest common divisor ([tex]$3$[/tex]):
[tex]$$
\frac{39}{51} = \frac{39 \div 3}{51 \div 3} = \frac{13}{17}.
$$[/tex]
So, the final answer is
[tex]$$
\frac{13}{17}.
$$[/tex]
[tex]$$
P(A \mid B) = \frac{P(A \cap B)}{P(B)},
$$[/tex]
where in this context:
- [tex]$A$[/tex] is the event "resident had a flu shot",
- [tex]$B$[/tex] is the event "resident is male."
From the table, we have:
- The number of male residents who had a flu shot is [tex]$39$[/tex].
- The total number of male residents is [tex]$51$[/tex].
Thus, the probability that a male resident had a flu shot is
[tex]$$
P(\text{Flu Shot} \mid \text{Male}) = \frac{39}{51}.
$$[/tex]
We can simplify the fraction by dividing the numerator and the denominator by their greatest common divisor ([tex]$3$[/tex]):
[tex]$$
\frac{39}{51} = \frac{39 \div 3}{51 \div 3} = \frac{13}{17}.
$$[/tex]
So, the final answer is
[tex]$$
\frac{13}{17}.
$$[/tex]
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