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- anonymous

In a certain part of the world there are more wet days than dry days. If a given day is wet, the probability that the following day will also be wet is 0.8 . If a given day is dry, the probability that the following day will also be dry is 0.6 . (I have answered the first few parts so I just write the one I do not know how to answer) In one season there were 44 cricket matches, each played over three consecutive days, in which the first and third days were dry. For how many of these matches would you expect that the second day was wet?

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- anonymous

- jamiebookeater

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- ybarrap

We know that a day is either wet or dry with probability 1:
$$
p(w)+p(d)=1
$$
We also know that a day is dry given the previous day was dry or wet given the previous day was dry with probability 1:
$$
p(d|d)+p(w|d)=1
$$
The probability that it rains today only depends on whether it rained yesterday, so the third day has no bearing on the problem.
We are looking for \(p(w)\), the probability that it rains on the second day. Once we have this, we get the average by multiplying the number of games played times the probability of a wet game: \(44\times p(w)\). This will give us the answer to the problem.
Let's find \(p(w)\).
We are given that \(p(w|w)=0.8\) and \(p(d|d)=0.6\)
A basic relation we can use is that
$$
p(w)=p(w\cap d)+p(w\cap w)=p(d)p(w|d)+p(w)p(w|w)
$$
We now know everything because
$$
p(d)=1-p(w)\\
p(w|d)=1-p(d|d)=1-0.6\\
p(w|w)=0.8
$$
One equation with one uknown - solvable.
Does this make sense?
Note, in this problem they give you that it was dry on day 1 and 3 but that doesn't matter because we just want the probability of rain today. We know the conditional probabilities, so we use this information alone to solve. If you didn't know the conditional probabilities, then you would need to estimate them with this sort of information, but that was not necessary here.

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