Question

the director of health services is concerned about a possible flu outbreak at her college. she surveyed 100 randomly selected residents from the college’s dormitories to see whether they had received a preventative flu shot. the results are shown below. what is the probability that a dormitory resident chosen at random from this group has had a flu shot, given that he is male?

	male	female	total
didnt have flue shot	12	8	20
total	51	49	100

residents at college dormitories

options: \\(\frac{51}{100}\\), \\(\frac{39}{51}\\), \\(\frac{39}{100}\\), \\(\frac{12}{17}\\)

Explanation:

Step1: Recall Conditional Probability Formula

The formula for conditional probability is \( P(A|B) = \frac{P(A \cap B)}{P(B)} \). In the context of this problem, let \( A \) be the event that a resident had a flu shot, and \( B \) be the event that a resident is male. So we need to find \( P(\text{Had Flu Shot} | \text{Male}) \).

Step2: Identify the Counts

From the table, the number of males who had a flu shot (which is \( n(A \cap B) \)) is 39, and the total number of males (which is \( n(B) \)) is 51.

Step3: Calculate the Probability

Using the formula for conditional probability with counts (since all residents are equally likely, we can use counts instead of probabilities), we have \( P(\text{Had Flu Shot} | \text{Male})=\frac{\text{Number of Males who had Flu Shot}}{\text{Total Number of Males}}=\frac{39}{51} \). We can simplify this fraction by dividing numerator and denominator by 3: \( \frac{39\div3}{51\div3}=\frac{13}{17} \). Wait, no, wait. Wait, the options have \( \frac{39}{51} \) simplified? Wait, no, let's check the options again. Wait, the options are \( \frac{51}{100} \), \( \frac{39}{51} \), \( \frac{39}{100} \), \( \frac{13}{17} \). Wait, 39 and 51: GCD of 39 and 51 is 3. 39 ÷ 3 = 13, 51 ÷ 3 = 17. So \( \frac{39}{51}=\frac{13}{17} \). Wait, but let's re - check the table. The total number of males is 51 (from the "Total" row and "Male" column), and the number of males who had a flu shot is 39 (from "Had Flu Shot" row and "Male" column). So the conditional probability is \( \frac{39}{51}=\frac{13}{17} \). Wait, but let's check the options. One of the options is \( \frac{13}{17} \). Wait, but let's do the calculation again.

Wait, the formula for conditional probability when dealing with a contingency table is: \( P(\text{Had Flu Shot} | \text{Male})=\frac{\text{Number of Males with Flu Shot}}{\text{Total Number of Males}} \). From the table, number of males with flu shot is 39, total number of males is 51. So \( \frac{39}{51}=\frac{13}{17} \) (dividing numerator and denominator by 3).

Answer:

\(\frac{13}{17}\) (or if we consider the non - simplified form, \(\frac{39}{51}\), but simplifying gives \(\frac{13}{17}\))