Need Correct Answer to Part 3 - Serap Umbrella Problem 4. Serap and her umbrella

Question

Need Correct Answer to Part 3 - Serap Umbrella

Problem 4. Serap and her umbrella 4.0/6.0 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? 0.67 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? No 3. Serap is carrying an umbrella and it is not raining. What is the probability that she saw the forecast? .05

Explanation / Answer

If Serap saw the forecast as "rain", then she will definitely carry an umbrella.

Since she is carrying an umbrella in this case, then this means she definitely saw the forecast as "rain".

Now, we need to find the probability that Serap saw the forcast as "rain" but actually it did not rain.

Using the information given to us,

P(forecast is "rain" AND it does not rain) = P(forecast is rain)*P(it does not rain | forecast is rain) = 0.2*0.2 = 0.04

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