Question

1. the high school students who attend springfield womens academy are eligible to tryout for various teams within the athletic department. students are 74% likely to tryout for softball; 27% likely to tryout for volleyball; 42% likely to tryout for tennis; and 88% likely to tryout for swimming. many students choose to tryout for multiple teams. students have equal probabilities of being freshmen, sophomores, juniors, or seniors. what is the probability of the union of trying out for either volleyball or swimming?

1.15
0.61
0.24
0.91

Explanation:

Step1: Recall the formula for the probability of the union of two events.

The formula for the probability of the union of two events \( A \) and \( B \) is \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). However, in this case, we are not given the probability of the intersection \( P(A \cap B) \). But we know that the maximum value of \( P(A \cap B) \) is the minimum of \( P(A) \) and \( P(B) \) (since the intersection can't be larger than either event). Here, \( P(\text{volleyball}) = 0.27 \) and \( P(\text{swimming}) = 0.88 \). The minimum of these two is \( 0.27 \), so the maximum possible value of \( P(A \cap B) \) is \( 0.27 \). But we can also note that if we assume independence (though the problem says many students choose multiple teams, but we can still use the formula with the given info, and since we need to find the union, we can use the formula as \( P(A \cup B)=P(A)+P(B)-P(A\cap B) \). But wait, actually, in probability, the union of two events \( A \) and \( B \) is at most \( 1 \), and at least \( \max(P(A), P(B)) \). Here, \( \max(0.27, 0.88)=0.88 \), and \( P(A)+P(B)=0.27 + 0.88 = 1.15 \), but since probability can't exceed \( 1 \), the intersection must be at least \( 0.27 + 0.88 - 1=0.15 \). But maybe the problem assumes that we can use the formula without the intersection (maybe it's a trick question where we consider that the intersection is \( 0 \), but that's not possible. Wait, no, actually, maybe the problem has a typo or we misread. Wait, no, let's check the options. The options are \( 1.15 \), \( 0.61 \), \( 0.24 \), \( 0.91 \). Wait, \( 0.27 + 0.88 = 1.15 \), but probability can't be more than \( 1 \), so that's impossible. Wait, maybe the question is actually using the formula where we assume that the intersection is \( 0 \), but that's wrong. Wait, no, maybe I made a mistake. Wait, the problem says "many students choose to tryout for multiple teams", but we are to find the probability of the union. Wait, maybe the problem is intended to use the formula \( P(A \cup B)=P(A)+P(B)-P(A\cap B) \), but since we don't have \( P(A\cap B) \), but we can see that the options include \( 0.91 \), which is \( 0.27 + 0.88 - 0.24 \)? Wait, no, \( 0.27 + 0.88 = 1.15 \), and \( 1.15 - 0.24 = 0.91 \). Wait, maybe the intersection is \( 0.24 \)? But how? Wait, maybe the problem is designed so that we use the formula and the correct answer is \( 0.91 \), because \( 0.27 + 0.88 - 0.24 = 0.91 \). Wait, but where does \( 0.24 \) come from? Wait, maybe the problem is actually a case where we use the formula and the intersection is \( 0.24 \), but maybe it's a mistake. Alternatively, maybe the problem is intended to use the formula as \( P(A \cup B)=P(A)+P(B)-P(A\cap B) \), and since the maximum possible union is \( 1 \), but the sum is \( 1.15 \), so the intersection must be at least \( 0.15 \), but the options include \( 0.91 \), which is \( 1.15 - 0.24 \), so maybe the intersection is \( 0.24 \). But let's check the options. The correct answer is \( 0.91 \), because \( 0.27 + 0.88 - 0.24 = 0.91 \). Wait, but how do we get \( 0.24 \)? Alternatively, maybe the problem is using the formula without considering the intersection, but that's wrong. Wait, no, maybe I misread the probabilities. Wait, the problem says: students are \( 74\% \) likely to tryout for softball; \( 27\% \) for volleyball; \( 42\% \) for tennis; \( 88\% \) for swimming. So \( P(\text{volleyball}) = 0.27 \), \( P(\text{swimming}) = 0.88 \). Then \( P(A \cup B)=0.27 + 0.88 - P(A \cap B) \). Now, since \( P(A \cap B) \) can be at most \( 0.27 \) (since \( 0.27 < 0.88 \)), so the minimum value of \( P(A \cup B) \) is \( 0.88 \) (when \( P(A \cap B)=0.27 \)), and the maximum is \( 0.27 + 0.88 = 1.15 \), but since probability can't exceed \( 1 \), the maximum is \( 1 \) (when \( P(A \cap B)=0.15 \)). But the options include \( 0.91 \), which is between \( 0.88 \) and \( 1 \). Let's check the options: \( 1.15 \) is too big (probability can't be more than \( 1 \)), \( 0.61 = 0.88 - 0.27 \), \( 0.24 = 0.88 - 0.64 \) (no), \( 0.91 \). Let's calculate \( 0.27 + 0.88 - 0.24 = 0.91 \). So maybe \( P(A \cap B)=0.24 \). So using the formula \( P(A \cup B)=0.27 + 0.88 - 0.24 = 0.91 \).

Step2: Calculate the probability.

Using the formula \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). We assume \( P(A \cap B) = 0.24 \) (maybe from the options, since \( 0.27 + 0.88 - 0.91 = 0.24 \)). So \( P(\text{volleyball} \cup \text{swimming}) = 0.27 + 0.88 - 0.24 = 0.91 \).

Answer:

\( 0.91 \) (the option corresponding to \( 0.91 \))