Problem 4. Serap and her umbrella 0.0/6.0 points (graded) Before leaving for wor
ID: 3070352 • Letter: P
Question
Problem 4. Serap and her umbrella 0.0/6.0 points (graded) Before leaving for work, Serap checks the weather report in order to decide whether to carry an umbrella. On any given day, with probability 0.2 the forecast is "rain" and with probability 0.8 the forecast is "no rain". If the forecast is "rain", the probability of actually having rain on that day is 0.8. On the other hand, if the forecast is "no rain', the probability of actually raining is 0.1. 1. One day, Serap missed the forecast and it rained. What is the probability that the forecast was "rain""? 2. Serap misses the morning forecast with probability 0.2 on any day in the year. If she misses the forecast, Serap will flip a fair coin to decide whether to carry an umbrella. (We assume that the result of the coin flip is independent from the forecast and the weather.) On any day she sees the forecast, if it says "rain she will always carry an umbrella, and if it says "no rain"" she will not carry an umbrella. Let U be the event that "Serap is carrying an umbrella", and let N be the event that the forecast is "no rain Are events U and N independent? Select an option 3. Serap is carrying an umbrella and it is not raining. What is the probability that she saw the forecast?Explanation / Answer
1.
P(rain) = P(forecast as rain and rain) + P(forecast as no rain and rain)
= (0.2*0.8 + 0.8*0.1)
P(forecast as rain/rain) = P(Forecast as rain and rain) / P(rain)
Thus, P(forecast as rain/rain) = 0.2*0.8 / (0.2*0.8 + 0.8*0.1)
= 0.16/0.24
= 2/3
= 0.67
2. option is not available with me
3, Serap carries umbrella and it is not raining. Serap was expecting rain
P(she saw forecast) = P(forecast as rain)
= P(forecast as rain and no rain) / {P(forecast as rain and no rain) + P(forecast as no rain and no rain)}
= 0.2*0.2 / {(0.2*0.2) + (0.8*0.9)}
= 0.04/0.76
= 1/19
= 0.0526
rounding off to 0.05
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