Question

Heavy Snowfall (H) 0.95 (Lines going to both Closed and Open). Light snowfall (L) 0.90 (Lines going both to Closed and Open). H (Heavy snowfall). L (Light snowfall). C (Closed). O (Open). The tree diagram relates the weather forecast and school closings. Find the probability (as a percent) that the school is closed tomorrow if there is a 65% chance for heavy snow.

Answer 1

To find the probability that the school is closed tomorrow given that there is a 65% chance of heavy snow, we can use the information given in combination with the law of total probability.

Let's define the events:
- \( H \): Heavy snowfall
- \( L \): Light snowfall
- \( C \): School is closed
- \( O \): School is open

The probabilities you provided are:
- \( P(C|H) = 0.95 \) (Probability of school being closed given heavy snowfall)
- \( P(O|H) = 0.05 \) (Probability of school being open given heavy snowfall, which is complementary to closed)
- \( P(C|L) = 0.90 \) (Probability of school being closed given light snowfall)
- \( P(O|L) = 0.10 \) (Probability of school being open given light snowfall, which is complementary to closed)

Given:
- \( P(H) = 0.65 \) (Probability of heavy snowfall)
- \( P(L) = 1 - P(H) = 0.35 \) (Probability of light snowfall)

Now, using the law of total probability to find \( P(C) \) (the overall probability that the school is closed):

\[
P(C) = P(C|H) \cdot P(H) + P(C|L) \cdot P(L)
\]

Substituting the values:

\[
P(C) = (0.95 \cdot 0.65) + (0.90 \cdot 0.35)
\]

Calculating each term:

\[
0.95 \cdot 0.65 = 0.6175
\]
\[
0.90 \cdot 0.35 = 0.315
\]

Now summing them up:

\[
P(C) = 0.6175 + 0.315 = 0.9325
\]

To convert this to a percentage:

\[
P(C) \times 100 = 93.25\%
\]

Thus, the probability that the school is closed tomorrow is approximately **93.25%**.