Conditional Probability and law of total probability

  • #1
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Homework Statement


It rains in a city with a chance of 0.4. The weather forecast is not always accurate. When there will be a rain the next day, the forecast predicts the rain with probability 0.8; When there is no rain, the forecast falsely predicts a rain with probability 0.1. You take your umbrella every time rain is forecast, and you take your umbrella 25% of the times when rain is not forecast. Find the chance that it actually rains given that the forecast predicts rain. Given that it rains, what is the probability that you do not have your umbrella?

Homework Equations


Conditional probability

The Attempt at a Solution


Let R be the event about rain tomorrow, FR be the event about weather forecast predicts will rain tomorrow and U about I bring my umbrella.

I got the first part, P(R | FR) = P(FR | R)P(R)/(P(FR | R)P(R)+P(FR | R^c)P(R^c))

But I don't know how to do the second part.
 

Answers and Replies

  • #2
PeroK
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Homework Statement


It rains in a city with a chance of 0.4. The weather forecast is not always accurate. When there will be a rain the next day, the forecast predicts the rain with probability 0.8; When there is no rain, the forecast falsely predicts a rain with probability 0.1. You take your umbrella every time rain is forecast, and you take your umbrella 25% of the times when rain is not forecast. Find the chance that it actually rains given that the forecast predicts rain. Given that it rains, what is the probability that you do not have your umbrella?

Homework Equations


Conditional probability

The Attempt at a Solution


Let R be the event about rain tomorrow, FR be the event about weather forecast predicts will rain tomorrow and U about I bring my umbrella.

I got the first part, P(R | FR) = P(FR | R)P(R)/(P(FR | R)P(R)+P(FR | R^c)P(R^c))

But I don't know how to do the second part.
For these problems I always use a probability tree. Its especially useful when it gets complicated by a third variable.
 
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For these problems I always use a probability tree. Its especially useful when it gets complicated by a third variable.
I don't see how to draw probability tree.
 
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PeroK
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I don't see how to draw probability tree.
Is that something you've never been taught?
 
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Is that something you've never been taught?
I never learn how to draw a tree
 
  • #6
PeroK
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I never learn how to draw a tree
Ok. I suggest you work out the 4 probabilities: FR+R, FR+NR, FNR+R, FNR+NR and see whether that gives you any ideas.
 
  • #7
Ray Vickson
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Homework Statement


It rains in a city with a chance of 0.4. The weather forecast is not always accurate. When there will be a rain the next day, the forecast predicts the rain with probability 0.8; When there is no rain, the forecast falsely predicts a rain with probability 0.1. You take your umbrella every time rain is forecast, and you take your umbrella 25% of the times when rain is not forecast. Find the chance that it actually rains given that the forecast predicts rain. Given that it rains, what is the probability that you do not have your umbrella?

Homework Equations


Conditional probability

The Attempt at a Solution


Let R be the event about rain tomorrow, FR be the event about weather forecast predicts will rain tomorrow and U about I bring my umbrella.

I got the first part, P(R | FR) = P(FR | R)P(R)/(P(FR | R)P(R)+P(FR | R^c)P(R^c))

But I don't know how to do the second part.

Consider, say, 1000 identical days. The given data implies that in 400 of those days it will rain (R), while 600 of those days will have no rain (N).

In the 400 rainy days, forecasted rain (FR) occurred on 80/% × 400 = 320 days, and forecasted no rain (FN) on 400-320 = 80 days.

In the 600 non-rainy days, forecasted rain occurred on 0.1 × 600 = 60 days and forecasted no-rain on 600-60 = 540 days.

To summarize:
$$\begin{array}{c|cc|c}
& \text{R} & \text{N} & \text{Total}\\ \hline
\text{FR} & 320 & 60 & 380 \\
\text{FN} & 80 & 540 & 620 \\ \hline
\text{Total} & 400 & 600 & 1000
\end{array}
$$
From this table you can easily read off the solution to part 1, and you should get the same result as you do from using Bayes.

To do the second part, look carefully at the stated assumptions about U (the number of umbrella days) in the different cells of the table. Do you think that enough information was given in the problem to allow you to fill in the precise U-amounts in both cells of row FN? Does the second part have a unique solution, or are a variety of answers possible while respecting the precise wording given in the problem? Is there a most plausible solution?

After doing part 2 using the above tabular method, it might be fun to translate all that back into Bayesian language.
 
Last edited:

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