To find the probability that the school is closed tomorrow given a 70% chance of heavy snow, we can layout the probabilities based on the provided tree diagram structure.
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Let's denote:
- \( P(H) \): Probability of Heavy Snow = 0.70
- \( P(L) \): Probability of Light Snow = 0.30 (since \( P(L) = 1 - P(H) = 1 - 0.70 \))
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The branches of the tree can be expressed in terms of the closure probabilities. You stated there is a 90% closure rate under light snow (\( P(C|L) \)) and a closure rate under heavy snow that is not explicitly given but is less than 95%.
For heavy snow, we will denote its closure rate as \( P(C|H) \) which is the probability of closure given heavy snow. Since we know \( P(C|H) < 0.95 \), let’s assume that \( P(C|H) = x \) where \( x \) is a probability less than 0.95.
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We can now calculate:
\[ P(C) = P(C|H) * P(H) + P(C|L) * P(L) \]
Plugging in the known values:
\[ P(C) = x * 0.70 + 0.90 * 0.30 \] \[ P(C) = 0.70x + 0.27 \]
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To express this as a percentage, multiply by 100:
\[ P(C) \times 100 = (0.70x + 0.27) \times 100 \] \[ P(C) \times 100 = 70x + 27 \]
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Since \( x < 0.95 \), let’s evaluate what \( P(C) \) could be at that upper limit:
If \( x = 0.95 \):
\[ P(C) = 0.70 * 0.95 + 0.27 = 0.665 + 0.27 = 0.935 \]
Thus,
\[ P(C) \times 100 = 93.5% \]
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Therefore, the probability that the school is closed tomorrow given the conditions is less than 93.5%. Since you want just the probability, I will not assume any specific value for \( x \).
So, the final answer, based on the information provided:
\[ P(C) \text{ is less than } 93.5% \]
Remember that this would need to meet your criteria of \( L < C < 0.95 \) as \( x \) would have to be chosen in line with that condition.
If you would like specific numbers or have further adjustments in conditions, more specific closure probabilities for heavy snow would help clarify the outcome.