Question

At a local school, some percentage of seniors attend prom. Only 61% of the students go to prom and the senior lock-in. The probability of a student attending senior lock-in given that they attended prom is 0.73. What percentage of seniors attend prom, rounded to the nearest whole percent?

None of the answers.
45%
84%
20%

Answer 1

To solve the problem, we can use the information provided about the students attending prom and the probability of seniors attending lock-in given that they attended prom.

Let:
- \( P(A) \) be the probability that a student attends prom (which is 0.61 or 61%),
- \( P(B|A) \) be the probability of attending senior lock-in given that they attended prom (which is 0.73),
- \( P(B) \) be the overall probability that a student attends senior lock-in.

We can use the law of total probability. The total percentage of seniors attending prom, denoted \( P(S) \), can be expressed in terms of the other probabilities.

Since we know that 61% of seniors go to prom and that 73% of those who go to prom also go to the lock-in, we can set up the equation using conditional probability. Let's denote the percentage of seniors who attend prom as \( P(S) \).

1. The percentage of students (including non-seniors) attending prom is:
\[
P(A) = P(S) \times P(A|S) + P(N) \times P(A|N)
\]
where \( P(A|S) \) is the probability that seniors attend prom and \( P(A|N) \) is the probability that non-seniors attend prom.

However, we do not have the values for \( P(A|N) \) or the percentage of non-seniors, so it's more straightforward to work directly with the information we have.

2. If \( P(S) \) is the percentage of seniors attending prom, then:
\[
\text{Total attending lock-in} = P(S) \times P(B|A)
\]

Since \( P(A) = 0.61 \) and \( P(B|A) = 0.73 \):

3. Using:
\[
P(B) = P(S) \times 0.73
\]
But we don't need to calculate \( P(B) \)—we need to isolate \( P(S) \).

Since \( 0.61 = P(S) \cdot P(A|S) + (1-P(S)) \cdot P(A|N) \)

Assuming \( P(A|S) = 1 \) (because if seniors are defined by prom attendance), we have:

4. If we take \( P(B) = 0.61 \) as an established fact for overall prom attendance, we can reason how much of this is attributed specifically to seniors.

So we can also determine that since 73% of those that attend prom (who also attend lock-in) applies:
\[
P(S) \cdot 0.73 \approx 0.61
\]

5. Thus you can solve for \( P(S) \),
\[
0.73P(S) = 0.61 \rightarrow P(S) = \frac{0.61}{0.73} \approx 0.8356164
\]
And converting to percentage, we get:
\[
P(S) \approx 83.56\%
\]

Rounded to the nearest whole percent,
\[
P(S) \approx 84\%
\]

So the answer is **84%**.