Marie is getting married tomorrow, at an outdoor

Ответ:

JsYTTiktok

29.01.2022

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.

$P(B|A) = \frac{P(B)*P(A|B)}{P(A)}$

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

$P(B) = \frac{5}{365} = 0.0137$

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

$P(B|A) = \frac{0.0137*0.9}{0.11096} = 0.1111$

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

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