Question

Dave fails quizzes with probability 1/4, independent of other quizzes.

What is the probability that Dave fails two quizzes in a row before he passes two quizzes in a row?

The solution starts with:

Let B be the event that Dave fails two quizzes in a row before he passes two quizzes in a row. Let us use F and S to indicate quizzes that he has failed or passed, respectively. We then have

P(B)=P(FF ∪SFF ∪FSFF ∪SFSFF ∪FSFSFF ∪SFSFSFF ∪···)

Why? I did not understand why P(B) is equal to the above probability.

Answer 1

Here, we are waiting for the end event passing two quizzes in a row. Also we want here that we should have at least one occurrence of FF before that.

The thing we dont want here is SS coming before FF.

P(B) = Probability that FF comes before SS, note that as there are infinite number of quizzes here, eventually both FF and SS are to occur sometime. This is also the reason that FF is contained in P(B).

Therefore in a way we are stopping here at FF but before SS happened.

therefore now the probability here is computed as:

Therefore 0.1269 is the required probability here.