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Marie is getting married tomorrow, at an outdoor ceremony in the desert. In recent years, it has rained only 5 days each year. Unfortunately, the weatherman has predicted rain for tomorrow. When it actually rains, the weatherman correctly forecasts rain 90% of the time. When it doesn't rain, he incorrectly forecasts rain 10% of the time. What is the probability that it will rain on the day of Marie's wedding, given the weatherman forecasts rain

User Rmag
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1 Answer

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Answer:

11.11% probability that it will rain on the day of Marie's wedding, given the weatherman forecasts rain

Explanation:

Bayes Theorem:

Two events, A and B.


P(B|A) = (P(B)*P(A|B))/(P(A))

In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.

In this question:

Event A: Forecast of rain.

Event B: Raining.

In recent years, it has rained only 5 days each year.

A year has 365 days. So


P(B) = (5)/(365) = 0.0137

When it actually rains, the weatherman correctly forecasts rain 90% of the time.

This means that
P(A|B) = 0.9

Probability of forecast of rain:

90% of 0.0137(forecast and rains)

10% of 1 - 0.0137 = 0.9863(forecast, but does not rain)


P(A) = 0.0137*0.9 + 0.9863*0.1 = 0.11096

What is the probability that it will rain on the day of Marie's wedding, given the weatherman forecasts rain


P(B|A) = (0.0137*0.9)/(0.11096) = 0.1111

11.11% probability that it will rain on the day of Marie's wedding, given the weatherman forecasts rain

User EddieDean
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