Answer:
11.11% probability that it will rain on the day of Marie's wedding, given the weatherman forecasts rain
Step-by-step explanation:
Bayes Theorem:
Two events, A and B.
[tex]P(B|A) = \frac{P(B)*P(A|B)}{P(A)}[/tex]
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
[tex]P(B) = \frac{5}{365} = 0.0137[/tex]
When it actually rains, the weatherman correctly forecasts rain 90% of the time.
This means that [tex]P(A|B) = 0.9[/tex]
Probability of forecast of rain:
90% of 0.0137(forecast and rains)
10% of 1 - 0.0137 = 0.9863(forecast, but does not rain)
[tex]P(A) = 0.0137*0.9 + 0.9863*0.1 = 0.11096[/tex]
What is the probability that it will rain on the day of Marie's wedding, given the weatherman forecasts rain
[tex]P(B|A) = \frac{0.0137*0.9}{0.11096} = 0.1111[/tex]
11.11% probability that it will rain on the day of Marie's wedding, given the weatherman forecasts rain