Will It Rain?

Probability Level 2

In general, John and Amy are right about the weather with probabilities 70% and 60%, respectively and independently. Given the conversation above, what is the probability it rains tomorrow in their home of Brilliantia?

\frac{41}{50}

\frac{9}{23}

\frac{22}{25}

\frac{14}{23}

\frac{13}{20}

\frac{14}{25}

6 solutions

Eli Ross Staff
Oct 11, 2015

There are two possible outcomes:

It rains, so John was right and Amy was wrong: $70\% \cdot 40 \% = 28\%.$

It doesn't rain, so John was wrong and Amy was right: $30\% \cdot 60\% = 18\%.$

Via Bayes' Theorem , the probability of rain is $\frac{28\%}{28\%+18\%} = \frac{28}{46} = \frac{14}{23}.$

I think this is right but please check. Thanks. I think that the probability of being right can be interpreted in two ways. One is, if it's indeed going to rain then the probability that john had predicted it right is 0.7 and that he had predicted that it is not going to rain is 0.3. The similiar thing for Amy. Assuming that this is what the question meant, we have to make one more assumption that the probability of raining on any day in brilliantata is 0.5 . Then we get the probability that it rains given that the conversation between John and Amy took place as shown is 14/23 as shown by your calculation. Other way to interpret being right is that when John predicts it's going to rain, then there is 0.7 probability that it indeed rains and 0.3 probability that it doesn't. Similiar for Amy. But in this assumption, I think that the predictions cannot be independent as against given in the question. So we cannot just multiply 0.7 and 0.3 to get 0.21 as answer.

Pranav Rao - 5 years, 4 months ago

I agree that the problem is not well - posed. Does each person predict every day? Do they independently, or do they always disagree (which might not be possible)? What are the probabilities of each predicting rain, and what is the probability of rain on a given date?

Will Heierman - 4 years, 4 months ago

What is the rational of multiplying 70% and 60% with 40% and 30% respectively??

Parth Samani - 4 years, 10 months ago

Most of the answers below as well as the "official" answer are not satisfying from the mathematical point of view. Let me try and give it a more mathematical treatment.

What is (not) known?

Let $X$ be the random variable for John, such that $X = 1$ means that he predicts rain and $X = 0$ that he predicts no rain. We shall write shorter $X_1$ for $X = 1$ and $X_0$ for $X = 0$ . In the same manner $Y$ for Amy and $Z$ for the actual rain. What we know from the statement of the problem is:

"John and Amy are right about the weather with probabilities $70\%$ and $60\%$ , respectively..." For the case of rain $Z_1$ and John $X_1$ this should read as $P[X_1 | Z_1] = 0.7$ , i.e. we know that the probability of John predicting rain given that it will actually rain is 0.7. In other words, if it rains, then John predicts rain $70 \%$ of the time. This can be confused with: "if John predicts rain then it will rain in 70% of the cases", i.e. $P[Z_1 | X_1] = 0.7$ . This is indeed very subtle and can be put for discussion! (The same holds for Amy as well as for the case of no rain.)
"John and Amy are right about the weather with probabilities $70\%$ and $60\%$ ... independently ". This is also a very subtle part. It should actually translate to $X$ and $Y$ are conditionally independent given $Z$ ! Formally this means $P[X|Y;Z] = P[X|Z]$ and $P[Y|X;Z] = P[Y|Z]$ . In other words, the probability of John being right about his prediction given a rain event and Amy's prediction is same as the probability of John being right about his prediction given only a rain event.
The probability of rain $P[Z_1]$ is indeed not given. It should actually say $P[Z_1] = \frac{1}{2}$ .

What is the question?

The question of the problem should be stated as: What is $P[Z_1 | X_1 ; Y_0]$ ? In other words, what is the probability that it will rain under the condition that John predicted rain and Amy predicted no rain?

Given all this information a solution can be presented.

What is the answer?

Using Bayes one gets:

$P[Z_1 | X_1 ; Y_0] = \dfrac{P[X_1 ; Y_0 | Z_1]\cdot P[Z_1]}{P[X_1 ; Y_0 ]} = \dfrac{ P[X_1 ; Y_0 | Z_1]\cdot P[Z_1]}{ P[X_1 ; Y_0 | Z_1]\cdot P[Z_1] + P[X_1 ; Y_0 | Z_0]\cdot P[Z_0] }$

Fom the definition of conditional probability we know that $P[X_1 ; Y_0 ; Z_1] = P[X_1 | Y_0 ; Z_1] \cdot P[Y_0 | Z_1] \cdot P[Z_1],$ which implies that

$P[X_1 ; Y_0 | Z_1] = \dfrac{P[X_1 ; Y_0 ; Z_1] }{P[Z_1]} = P[X_1 | Y_0 ; Z_1]\cdot P[Y_0 | Z_1].$

The same holds for the complement case $Z_0$ . The assumption about conditional independence of $X$ and $Y$ given $Z$ reads as:

$P[X_1 | Y_0 ; Z_1] = P[X_1 | Z_1]$ and $P[X_1 | Y_0 ; Z_0] = P[X_1 | Z_0]$ .

In this case one gets

$P[X_1 ; Y_0 | Z_1] = P[X_1 | Z_1]\cdot P[Y_0 | Z_1]$ and $P[X_1 ; Y_0 | Z_1] = P[X_1 | Z_0]\cdot P[Y_0 | Z_0]$ .

Using the Bayes formula above it follows that:

$P[Z_1 | X_1 ; Y_0] = \dfrac{P[X_1 | Z_1]\cdot P[Y_0 | Z_1]\cdot P[Z_1]}{P[X_1 | Z_1]\cdot P[Y_0 | Z_1]\cdot P[Z_1] + P[X_1 | Z_0]\cdot P[Y_0 | Z_0]\cdot P[Z_0] }$

In order for the "official" answer to hold the following is assumed:

The probability of rain $P[Z_1] = P[Z_0] = \frac{1}{2}$ ;
conditional independence $P[X_1 | Y_0 ; Z_1] = P[X_1 | Z_1]$ and $P[X_1 | Y_0 ; Z_0] = P[X_1 | Z_0].$

I am not really sure if 2. immediately follows for everybody from the formulation of the original question. 1. is on the other side nowhere stated. However, I might be missing something ;)

Cheers

Johnny Micic - 4 years, 9 months ago

Mathematically satisfying

Ahmed Hossain - 2 years, 11 months ago

The sum of all the probabilities should be 100%, but 28%+18%=46%, the 2 other possibilities cannot be right... So what are the 54%? Hail? Snow?

Marc BAJET - 4 years, 10 months ago

The other two possibilities are "both right" and "both wrong".

Richard Desper - 4 years, 1 month ago

Marc Bajet, the reason it doesn't sum to 100% is that the two other outcomes have been excluded: that both are right, or both are wrong.

Uffe Jensen - 4 years, 10 months ago

Exactly @Parth Samani

Varun Nagraj - 4 years, 9 months ago

I don't get it

Vlad Opriş - 4 years, 9 months ago

Bayes' theorem cannot be applied here because we do not know if it rained the next day, or at least the probability that it will rain on any day is given in which case we can assume that it rained with that probability. and so it is independent of what anyone said about if it will rain or not.

suresh apte - 4 years, 9 months ago

My intuition brought me to the same/correct answer. But could anyone tell me what's wrong with the following approach? P(John is right or Amy is wrong) = P(John is right and Amy is wrong). Using the inclusion-exclusion principle, P(John is right and Amy is wrong) = P(John is right or Amy is wrong) = P(John is right) + P(Amy is wrong) - P(John is right and Amy is wrong). It follows that P(it rains) = P(John is right and Amy is wrong) = 0.55.

James Wilson - 3 years ago

A key is that the predictions are independent. That is a very strange property for weather forecasters! That unrealistic aspect makes the question less enjoyable.

I found it useful to reframe the questions as, "What is the chance that John is correct given that they disagree?" Since their predictions are independent, we can work out that they disagree a surprising 46% of the time! When they are disagreeing, John is the better bet, 28/46 - as shown in Ross' answer.

Adam Hammond - 2 years, 10 months ago

You are assuming that there is an a priori chance of 50% for rain on any given day.

You need to explicitly mention this prior.

Bahman Engheta - 2 years, 9 months ago

Laurent Shorts
Apr 19, 2016

The probability is $\frac{14p}{5p+9}$ , where $p$ is the probability to have rain on any random day. So we cannot answer, unless we suppose $p=0.5$ and therefore the answer would be $\frac{14·\frac{1}{2}}{5·\frac{1}{2}+9}=\frac{14}{23}$ , but this should be clearly stated .

For example, if in Brilliantia there's always sunshine, the answer would be 0, (whatever they both would be saying).

Yes, that is what I got: Using Bayes' Theorem, we would have

R: it rains, $\not R$ : it won't rain

A: John says it will rain and Amy says it will not.

Then $p(R | A) = \frac{p(A | R) p(R) }{ p(A | R) p(R) + p(A | \not R) p(\not R) }$

Obviously, the stated solution is only correct if the probabilities for R and $\not R$ are the same.

Susanne Yelin - 5 years ago

Yes, indeed. I tried to explain this in a comment to the official solution. There is also a sublte point on the independence of the John's and Amy's predictions. Here I shall try to present it in your notation: $A$ = John says it will rain, $B$ = Amy says it will rain, $R$ it rains. $A^c$ , $B^c$ , $R^c$ for the opposite events. Then

$P[R | A ; B^c] = \dfrac{ P[A ; B^c | R]\cdot P[R]}{ P[A ; B^c | R]\cdot P[R] + P[A ; B^c | R^c ] \cdot P[ R^c] }.$

The question is, how do we compute $P[A ; B^c | R]$ and $P[A ; B^c | R^c]$ ? From the definition of conditional probability we know that $P[A ; B^c | R] = P[A | B^c ; R] \cdot P[ B^c | R]$ (and the same for $R^c$ . ) If we assume that $A$ and $B^c$ are conditionally independent given $R$ , i.e. $P[A | B^c ; R] = P[A | R]$ , we would have that $P[A ; B^c | R] = P[A | R] \cdot P[ B^c | R] = 0.7 \cdot 0.4$ (and same for the oposite event of not raining). Hence the suggested solution would follow.

It is stated in the original formulation that John and Amy are right about the weather independently. This should probably read as $A$ and $B$ are conditionally independent given $R$ . However I am not sure if everybody reads it that way.

Cheers.

Johnny Micic - 4 years, 9 months ago

There is some information though, about the weather, because Amy is right 60% and thinks it might not be sunny. Meta information. Dunno what the ramifications are.

Michael Stevenson - 3 years, 8 months ago

You will need the prior. Assuming prior = 0.5 you get the solution

Gang Su - 3 years, 6 months ago

You don't need to know the true probability of rain. In Ross's explanation and the identical technique I used to solve the problem, the probabilities of accuracy are addressed. This essentially substitutes the need to know what the actual chance of rain is if the forecast's percentages of true accuracy thereof are given.

Marc Moncada - 3 years ago

Does the probability of rain make a difference? And if so, can I understand why without resorting to Bayes Theorem? Yes and yes.

My approach to answering this was to imagine that it rained on 100 separate occasions. On 28 such occasions John would predict rain and Amy wouldn't. Then I imagined 100 days without rain and worked out that on 18 such occasions John would predict rain and Amy wouldn't. So over those 200 days only 46 would satisfy the given conditions (John yes; Amy no) and 28 out of the 46 would be rainy days.

In reading Laurent's comments I had to think where I had gone wrong. It was in the last sentence. Why expect 200 days in which it rained exactly half the time? What if it only rained a quarter of the time? Then I would need to consider 400 days of which 100 were rainy, giving 28+18+18+18 = 82 as the number of days satisfying the given condition and the probability of rain being 28 out of 82.

Paul Cockburn - 2 years, 6 months ago

We need a prior. Without any idea, we can suppose that any possibility has equal chance to happen (bad, but better than nothing).

But here, that would not mean we can assume there's a 50% chance of rain and 50% of sun every day. $\dfrac{14}{23}=0.609$ would be the answer if you were in a land where it's always sunny, or always rainy, and you'd assume there's a 50% probability for each case.

The best way to do it — that is, without knowing anything about the average probability of rain in the place where they live — would be to suppose each probability $p$ of rain equally possible.

The answer would then be: $\displaystyle{\int_0^1\!\!\!\dfrac{14p}{5p+9}\text{d}p\ =\ \dfrac{14}{5}-\dfrac{126}{25}\ln\Big(\frac{14}{9}\Big)\ =\ 0.573}$ .

Laurent Shorts - 2 years, 6 months ago

Sujith Kumar
Oct 1, 2015

P (A intersection B ) =P (A)×P (B) = 0.7×0.6 = 0.42 = 14/25

Not exactly (the answer is 14/23). Is it possible for them to both be correct in this situation?

Eli Ross Staff - 5 years, 8 months ago

I just want to add up few facts here. For the rain to occur, John has to be true and Amy has to be false. And because it is a AND situation. We need to multipy their respective probabilities. P(J) =7/10 P(A')=1-(6/10)=4/10

Req prob=(7/10)*(4/10)=28/100=14/50=7/25

PS: I know this may be false for that option didn't exist. Pls crt me where I went wrong. TY!!

Rajeshwar Gupta - 5 years, 8 months ago

There are four cases, whose probabilities add up to 1:
John right, Amy right
John right, Amy wrong
John wrong, Amy right
John wrong, Amy wrong

Note, however, that only two of these cases are even possible. How does this affect things?

Eli Ross Staff - 5 years, 8 months ago

@Sujith Kumar , @Rajeshwar Gupta , Laurent Shorts' explanation captures the issue correctly. Read his.

But this might help: you are thinking of the inverse problem. You are solving for "what is the probability that John is right & Amy is wrong. In that case, your calculations are exactly correct.

However, the problem is asking for the inverse: what is the probability that it will rain , assuming that John has claimed it will, and Amy has claimed it won't.

This requires "inverting" the priors, such that "probability of A given B" becomes "probability of B given A". That uses Bayes' Theorem, exactly as Laurent described in his solution.

Note also that the "official" answer is wrong. Or, not wrong, per se, but assuming that the likelihood of rain is 50/50.

Mike Williamson - 4 years, 10 months ago

But the given question has ans, 14/23

Darshan Baid - 5 years, 8 months ago

ARE U JOKING ME? 0.42 DOES NOT EQUAL 14/25

William G. - 4 years, 9 months ago

Robert Williams
Jan 2, 2018

Representing the possibilities in a table:

	Amy Wrong 40%	Amy Right 60%
John Wrong 30%	Not possible	18%
John Right 70%	28%	Not possible

So when they disagree John will be right in the ratio 28:18

$\frac{28}{46} = \frac{14}{23}$

N B
Aug 18, 2016

I got the correct answer but I don't feel like I quite understand it, so if anyone here could help make my reasoning more intuitive, that would be greatly appreciated.

Naively, the probability of it raining is the probability that John is right and Amy is wrong:

$P(\text{rain}) = \frac{7}{10} \times \frac{4}{10}$

However, this is incorrect because the probability of it not raining by this reasoning is the probability that John is wrong and Amy is right:

$P(\text{no rain}) = \frac{3}{10} \times \frac{6}{10}$

and

$\frac{7}{10} \times \frac{4}{10} + \frac{3}{10} \times \frac{6}{10} \neq 1$

So we must normalize our probabilities by

$\frac{1}{\frac{7}{10} \times \frac{4}{10} + \frac{3}{10} \times \frac{6}{10}} = \frac{50}{23}$

and therefore,

$P(\text{rain}) = \frac{7}{10} \times \frac{4}{10} \times \frac{50}{23} = \frac{14}{23}$

I don't really understand why these steps are justified, I just know that they work. Perhaps this is because I have no formal experience with probability theory, just indirect exposure in physics classes. I really like explanations that involve decision-tree diagrams and elementary things like, ``X fraction of the balls are black, Y fraction are white, blah blah insert intuitive reasoning here", so if anyone has an explanation like that, that would be greatly appreciated!

Bruce Bartholomew
Sep 4, 2016

It rains every day in Brilliantia. John and Amy are both very poor weather predictors, so they haven't noticed this. John predicts rain 70% of the time, while Amy predicts rain 60% of the time. John has predicted it will rain tomorrow and Amy has predicted a dry day.

I have met the conditions of the problem, and yet the probability it will rain tomorrow is 100%.

To arrive at the given solution, one must make another assumption, that John and Amy are 70% and 60%, correct, respectively, in their predictions of rain as well as in their predictions of dry days. Without the assumption, the answer cannot be determined.

I don't know if your right, but I like it.

Bridget Collier - 3 years, 5 months ago

Will It Rain?

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