This is Problem 1.39 from Tsitsiklis, Bertsekas, Introduction to Probability, 2nd edition.
My answer was $P(A) = W (P_g * k) + W^c (P_b * k)$
where W is the probability of good weather and k is the number of students required for the class to be taught.
The solution book says it is
My thinking is that:
-Each student chooses independently whether to go to class
-They have $P_g$ given good weather and $P_b$ given bad weather
-The multiplication rule says you can calculate many independent events by the probability of each event * the number of events
Why doesn't the multiplication rule apply?
Best Answer
A bit incorrect.
You wrote $$P(A)=W(P_g*k)+W^c(P_b*k)$$ But, since event that one particular student turns up is independent of other, you have a probability of $P_g$ or $P_b$ for each one of those k students; so, it is not $P_g*k$, but rather, $P_g^k$ and $P_b^k$.
So, it would reduce to $$P(A)=W(P_g^k)+W^c(P_b^k)$$
But wait. We need to select k students out of n and then find total probability, which will be:
P(prof teaches)=P(teaches when weather is good) + P(teaches when weather is bad)
Hint:
Do you know conditional probability?
Use $$P(teaches) = P(bad)P(teaches/bad) + P(good)P(teaches/good)$$ That summation shown in P(A/B) and P(A/$B^c$) step is used to include all possibilities of no. of students turning up for a given weather, whose iteration begins from i=k because prof needs at least k students out of n for lecture.