An Application of Bayes' Rule

I am working on the following problem:

A school serves orange juice on Thursdays and milk every other day. It is warm outside 70% of the time, Harry is only happy when orange juice is served or it is warm outside. Harry is happy today. What is the probability today is a warm Thursday?

I think this might be an application of Bayes' theorem? (I could be wrong) But I cannot get beyond formulating the following: $$P(\text{Warm Thursday} | \text{Happy}) = \frac{P(\text{Happy} | \text{Warm Thursday})P(\text{Warm Thursday})}{P(\text{Happy})}$$

I believe $$P(Warm) = 0.7$$ but I don't know where to go from here working out the rest, any help would be great, thanks.

• This is somewhat poorly worded. How many days are in the school week? Are we to assume that "warm" and "thursday" are independent? Are we to assume that Harry is $\textit {always}$ happy on warm days or when he gets orange juice? (easily pleased fellow, that Harry). Assuming all those points are sorted out, just proceed by cases. What's the probability that it is a warm Thursday? What's the probability that it is a cold Thursday? What's the probability that it is a warm non-Thursday?
– lulu
Dec 4, 2019 at 1:31
• I certainly agree, it is poorly worded - but it's all I have. I think its safe to assume the points you raised are sorted out, I am just considering the situation that there are 5 days in a school week and yes, Harry is $\textit{always}$ happy on warm days or when he gets orange juice, since there is no other information given about this. I don't understand how to work out the probability for the cases?
– gb4
Dec 4, 2019 at 1:40
• So, under those assumptions, the probability that it is a warm thursday is the product of the probability that it is warm and that it is thursday. That's what independent means. (and I guess we should assume that it is a school day). So I'd say that was $.7\times .2$ (different assumptions might lead to different values). The other cases are similar.
– lulu
Dec 4, 2019 at 1:43
• So, the probability of it being a warm Thursday under this assumption is just $0.7 \cdot \frac{1}{5} = 0.14$. Maybe the assumption on independence is too strong then, since most of the information in the question is not even used if this is the case?
– gb4
Dec 4, 2019 at 1:47
• That wasn't what you were asked. You were asked for the probability that it was a warm thursday $\textit {given}$ that Harry was happy. You have to work out a couple more cases and then divide. the conditional probability that you wrote is correct. You have computed the numerator, now you need the denominator.
– lulu
Dec 4, 2019 at 1:49

I'm assuming that we're talking about the whole week (7 days), not the school week (5 days). Assuming independence: $$P(\text{Warm Thursday}) = P(\text{Warm}) \times P(\text{Thursday}) = 7/10 \times 1/7 = 0.1$$ Then, assuming Harry is always happy when it's warm: $$P(\text{Happy} | \text{Warm Thursday}) = 1$$ Finally, what's the prior probability of him being happy? Use marginalization $$P(\text{Happy}) = P(\text{Happy on Monday}) + P(\text{Happy on Tuesday}) \ + \ ...= 1/7 + 6 \times 1/7 \times 0.7$$

Note that the probability of being happy on Thursday is the probability of Thursday itself since we know he's happy on Thursday.

Use Bayes Rule to combine the results.

Alternatively, you can do it this way. Being happy is the same as either warm or Thursday. It implies the probability of being happy is given by

$$\operatorname{Pr}( \text{Happy}) = \operatorname{Pr}( \text{Warm} \cup \text{Thursday})$$

Then we can use the inclusion-exclusion principle

$$\operatorname{Pr}( \text{Warm} \cup \text{Thursday}) = \operatorname{Pr}( \text{Warm}) + \operatorname{Pr}( \text{Thursday} ) - \operatorname{Pr}( \text{Warm} \cap \text{Thursday})$$

We know that $$\operatorname{Pr}( \text{Warm} \cap \text{Thursday}) = 70\% \times \frac{1}{7} = 10\%$$.

Then

$$\operatorname{Pr}( \text{Happy}) = 70\% + \frac{1}{7} - 10\%$$

You can finish the rest, as the other two quantities in the Bayes' rule formula are given by Tomasz Bartkowiak. The answer is the same as the answer given by Tomasz. I'm just computing the denominator in the Bayes' rule formula in a different way.