Before leaving for work, Victor checks the weather report in order to decide whether to carry an umbrella. The forecast is “rain" with probability 20% and “no rain" with probability 80%. If the forecast is “rain", the probability of actually having rain on that day is 80%. On the other hand, if the forecast is “no rain", the probability of actually raining is 10%.
1. One day, Victor missed the forecast and it rained. What is the probability that the forecast was “rain"?
2. Victor misses the morning forecast with probability 0.2 on any day in the year. If he misses the forecast, Victor 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 he sees the forecast, if it says “rain" he will always carry an umbrella, and if it says “no rain" he will not carry an umbrella. Let U be the event that “Victor is carrying an umbrella", and let N be the event that the forecast is “no rain". Are events U and N independent?
3. Victor is carrying an umbrella and it is not raining. What is the probability that he saw the forecast?
2) NO
1. 0.6667
3.0.2963
1/2
1. To determine the probability that the forecast was "rain" given that it actually rained, we can use Bayes' theorem. Let's denote the events as follows: F1 = forecast is "rain" and R = it rained. We are looking for the probability of F1 given R, i.e., P(F1|R).
Using Bayes' theorem, we have:
P(F1|R) = (P(R|F1) * P(F1)) / P(R)
Given in the problem statement:
P(F1) = 0.2 (probability of the forecast being "rain")
P(F2) = 0.8 (probability of the forecast being "no rain")
P(R|F1) = 0.8 (probability of rain given the forecast is "rain")
P(R|F2) = 0.1 (probability of rain given the forecast is "no rain")
To calculate P(R), we need to consider the total probability of it raining, regardless of the forecast:
P(R) = P(R|F1) * P(F1) + P(R|F2) * P(F2)
= 0.8 * 0.2 + 0.1 * 0.8
Now we can substitute these values into Bayes' theorem to find P(F1|R).
2. To determine if events U (Victor carrying an umbrella) and N (forecast is "no rain") are independent, we need to compare the joint probability of U and N with the product of their individual probabilities.
If U and N are independent, then P(U ∩ N) = P(U) * P(N).
Given in the problem statement:
P(U|F1) = 1 (Probability of carrying an umbrella given the forecast is "rain")
P(U|F2) = 0 (Probability of carrying an umbrella given the forecast is "no rain")
P(N) = 0.8 (probability of the forecast being "no rain")
Now we can calculate P(U ∩ N) and compare it with P(U) * P(N) to determine if events U and N are independent.
3. To determine the probability that Victor saw the forecast, given that he is carrying an umbrella and it's not raining, we can use Bayes' theorem. Let's denote the events as follows: S = Victor saw the forecast, U = carrying an umbrella, and ¬R = it's not raining.
We want to find P(S|U and ¬R), i.e., the probability of seeing the forecast given that Victor is carrying an umbrella and it's not raining.
Using Bayes' theorem, we have:
P(S|U and ¬R) = (P(U and ¬R|S) * P(S)) / P(U and ¬R)
To calculate P(U and ¬R), we need to consider the total probability of Victor carrying an umbrella and it's not raining, regardless of seeing the forecast:
P(U and ¬R) = P(U and ¬R|S) * P(S) + P(U and ¬R|¬S) * P(¬S)
We are given the following probabilities:
P(U|F1) = 1 (Probability of carrying an umbrella given the forecast is "rain")
P(U|F2) = 0.5 (Probability of carrying an umbrella given the forecast is "no rain")
P(¬R|F1) = 0.2 (Probability of it not raining given the forecast is "rain")
P(¬R|F2) = 0.9 (Probability of it not raining given the forecast is "no rain")
P(S) = 1 - P(missing the forecast) = 1 - 0.2 = 0.8
Now we can substitute these values into Bayes' theorem to find P(S|U and ¬R).