Before leaving for work, Victor 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.

One day, Victor missed the forecast and it rained. What is the probability that the forecast was “rain"?

- unanswered

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?

- unanswered

Victor is carrying an umbrella and it is not raining. What is the probability that he saw the forecast?

1. 2/3

2. No

3. 0.2963

how to calculate question 3?

Ah, the joys of probability! Let's tackle this question with a dash of humor, shall we?

Well, if Victor is carrying an umbrella and it's not raining, we can safely assume that Victor might have believed the forecast that said "no rain." But how confident can we be? To find out the probability that he saw the forecast, we need to calculate the conditional probability, given the information we have.

But before we dive into the calculations, let me ask you this: Did Victor get his umbrella from a clown? Because that would explain a lot about his decision-making process!

Anyway, back to the probabilities. We know that the probability of carrying an umbrella given that the forecast is "no rain" is 0. On the other hand, if the forecast is "rain," the probability of carrying an umbrella is 1. Given that it's not raining, the probability of the forecast being "no rain" increases. So, it's safe to say that the probability of Victor having seen the forecast is pretty high.

But remember, life is full of surprises, just like a clown pulling a rubber chicken out of a hat. So, while it's likely that Victor saw the forecast, there's always a chance that he just has an irrational fear of getting wet and decided to carry the umbrella anyway. I mean, who knows what goes on in Victor's mind?

In conclusion, the probability that he saw the forecast is high, but as high as the probability of a clown getting a standing ovation for their jokes? Well, that's up for debate!

Nah they have a clown bot?

To answer the question, we need to consider the given probabilities and use Bayes' theorem. Bayes' theorem states that:

P(A|B) = (P(B|A) * P(A)) / P(B)

In this case, we want to find the probability that Victor saw the forecast given that he is carrying an umbrella and it is not raining. We can denote the events as follows:
A = Victor saw the forecast
B = Victor is carrying an umbrella and it is not raining

We are looking for P(A|B), which is the probability of A given B.

From the given information, we can determine:
P(A) = probability of seeing the forecast = 1 - probability of missing the forecast = 1 - 0.2 = 0.8
P(B|A) = probability of carrying an umbrella and it is not raining given that Victor saw the forecast = 0 (since if he saw the forecast and it says "no rain", he will not carry an umbrella)
P(B) = probability of carrying an umbrella and it is not raining

To find P(B), we can use the law of total probability. We consider two cases:
1. Victor saw the forecast and it says "rain"
2. Victor saw the forecast and it says "no rain"

For case 1, the probability of carrying an umbrella and it is not raining is 0 (since if he saw the forecast and it says "rain", he will always carry an umbrella).

For case 2, the probability of carrying an umbrella and it is not raining is the probability of carrying an umbrella and it is not raining given that Victor saw the forecast says "no rain". From the given information, this probability is 0.9 (1 - 0.1, since the probability of actually raining given the forecast says "no rain" is 0.1).

Since these two cases are mutually exclusive, we can calculate P(B) as the sum of their probabilities:
P(B) = P(saw forecast and it says "rain") * P(carrying an umbrella and it is not raining in that case) + P(saw forecast and it says "no rain") * P(carrying an umbrella and it is not raining in that case)
P(B) = 0 * 0 + 0.8 * 0.9 = 0.72

Now, we can substitute these values into Bayes' theorem:
P(A|B) = (P(B|A) * P(A)) / P(B)
P(A|B) = (0 * 0.8) / 0.72
P(A|B) = 0

Therefore, the probability that Victor saw the forecast given that he is carrying an umbrella and it is not raining is 0.