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The probability it rains on Wednesday this week is 40%, while the probability it rains on Thursday this week is 30%. However, it is twice more likely to also rain on Thursday, if it has already rained on Wednesday. What is the probability it rains at least one of the two days?

Although I am not familiar with probabilities (I just did a bit of reading), I will try to start: The probability it rains at least one of the two days is 1- the probability it will not rain on any of the two days. This is 1-(1-40%)(1-30%)=58%

But this is only when the two events are independent. The probability it rains on Thursday is not independent. How do I take into account this?

Thanks for your help.

Marius Stephant
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    Since there is a question of what "twice more likely" means in the answers below, it requires some clarification. A Bayesian interpretation would mean "twice the odds ratio", but another interpretation might be "twice the probability". Can you clarify by looking in the book or asking the instructor or even giving the expected answer? – robjohn Jun 17 '18 at 13:11

5 Answers5

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The point in Probabilities is the following: Denote $W$ the event that it rains on Wednesday, ${\bar W}$ the event that it doesn't. Likewise with $T$ and $\bar T$ for Thursday.

Then $P(W) =0.4$.

By marginalization, you have $0.3 = P(T) = P(T|W) P(W) + P(T|{\bar W}) P({\bar W}) $. So if you are interested in conditional events like $P(T|W)$, you could use this formula as follows. The additional info is that $ P(T|W) = 2 P(T|{\bar W}) $. So you obtain $0.3 = P(T|W) 0.4 + \frac12 P(T|W) 0.6$. This allows you to calculate $P(T|W) = 3/7$.

Now the situation $P*$ that it rains on any of the two days can be split into two disjoint events, so

$P* = P(W) + P(T|{\bar W}) P({\bar W}) = P(W) + \frac12 P(T|W) P({\bar W})= 0.4 + \frac12 \frac37 0.6 = \frac{37}{70} \simeq 0.528$

Sasho Nikolov
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Andreas
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  • But then we take P(T|W) = 3/7 and P(W) = 4/10, so the requested is 1-(1-4/10)(1-3/7) = 65.71%. No? – Marius Stephant Jun 17 '18 at 10:15
  • That is again for independent events. I modified the answer. – Andreas Jun 17 '18 at 10:17
  • Thank you. But why do you add the two? – Marius Stephant Jun 17 '18 at 10:17
  • I added disjoint events: It either rains on Wednesday, then we are done with P* (it has rained on any of the two days). Or it doesn't rain on Wednesday, then the only event which leads to P* is that it also rains on Thursday. -- And I initially forgot the $P({\bar W})$, sorry. – Andreas Jun 17 '18 at 16:29
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$p$ is the probability that it rains on Thursday if it has rained on Wednesday. $q$ is the probability that it rains on Thursday if it has not rained on Wednesday.

Given that the probability it rains on Thursday is $30\%$, we have $$ \frac25p+\frac35q=\frac3{10}\tag1 $$ Given that it is twice as likely to rain on Thursday if it has rained on Wednesday, we have $$ \frac{p}{1-p}=2\frac{q}{1-q}\tag2 $$ Solving $(1)$ and $(2)$ simultaneously, we get $$ p=\frac{17-\sqrt{193}}{8},q=\frac{-11+\sqrt{193}}{12}\tag3 $$ The probability that it rains on at least one day is the complement of the probability that it doesn't rain on either day. That is, $$ 1-\frac35(1-q)=\frac{-3+\sqrt{193}}{20}\approx54.4622\%\tag4 $$

robjohn
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  • Where does (2) come from? Why isn't it simply $p=2q$ (as the other answers have)? – Especially Lime Jun 17 '18 at 11:52
  • "Twice as likely" is not "twice the probability". See this answer for an example. – robjohn Jun 17 '18 at 11:55
  • I'd be surprised if your interpretation ("likelihood" reflecting odds rather than probability) was what was meant here. It certainly does not seem to be standard mathematical usage, see e.g. https://en.wikipedia.org/wiki/Likelihood_function . – Especially Lime Jun 17 '18 at 12:18
  • In What does "twice as likely" mean?, there is no consensus, but most answers suggest the odds ratio is doubled. – robjohn Jun 17 '18 at 12:26
  • That question explicitly says that what people normally mean is "twice as probable", but goes on to ask if there's a way to define "likely" with different properties. There is no suggestion that such a definition is standard (in fact the top-voted answer says that it is not). – Especially Lime Jun 17 '18 at 12:36
  • In particular, the title is not an accurate summary of the question. – Especially Lime Jun 17 '18 at 12:38
  • Actually, Henning Makholm's answer does agree, but uses the log of the odds ratio and adds $\log(2)$. – robjohn Jun 17 '18 at 12:56
  • However, without further clarification, it may well be that twice the probability is what was intended in this problem. It definitely wasn't in the question to my answer that I cited above. I have commented on the question asking for clarification. – robjohn Jun 17 '18 at 13:05
  • Henning's answer says that he doesn't think "twice as likely" as something with probability greater than $1/2$ should be meaningful. He goes on to say that if you do want it to be meaningful, using the odds ratio is the best way to do that. So it sounds to me like he thinks the normal meaning is "double the probability". – Especially Lime Jun 17 '18 at 13:55
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P(T)=0.3=P(W)*P(T|W)+(1-P(W))*P(T|W)/2=0.4*P(T|W)+0.6*P(T|W)/2=0.7*P(T|W)

P(T|W)=0.3/0.7

P(T|$\lnot$W)=P(T|W)/2=3/14

P($\lnot$W$\land$$\lnot$T)=P($\lnot$W)(1-P(T|$\lnot$W))=0.6(1-3/14)=33/70

Djura Marinkov
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The full joint distribution of two Bernouilli random variables $W$, $T$ is given by the four probabilities $P(W, T), P(W, \overline T), P(\overline W, \overline T), P(\overline W, \overline T)$. These four numbers always satisfy the linear equation:

$$P(W, T)+P(W, \overline T)+P(\overline W, \overline T)+P(\overline W, T)=1$$

You've been given three other linear equations:

$$P(W,T)+P(W,\overline T)=0.4$$ $$P(W,T)+P(\overline W,T)=0.3$$ $$P(W,T)=0.24$$

We get the last equation from knowing that $P(T|W)=0.6$, $P(W)=0.4$ and the identity $P(A,B)=P(A|B)P(B)$.

You now have four equations for four unknowns, which can be solved for not just the answer to the question, but all possible information about the events $W$ and $T$ and the relationships between them.

Jack M
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Let's rephrase :)

The summer session has $70$ days. Over this span, $28$ exams are scheduled in the morning and $21$ in the afternoon. However, the professors prefer to schedule double-exam days rather than spreading their activity over more "simple" days.

Let

$p$ be the number of afternoon exams scheduled in the $28$ morning exam days, and

$q$ the number of exams scheduled in the afternoons of the other $42$ days.

We know that $p + q = 21$.

$p/28$ is the probability to have an afternoon exam after a morning exam,

$q/42$ is the probability to have an afternoon exam after a free morning.

So $p/28 = 2 \times q/42$ that gives $q = 9$

The total number of exam days is then $28 + 9 = 37$.

Boyku
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