figs wrote:can someone explain this exercise with another way. I'm interesting in know how to do with conditional probability.
Let E be the event that the first 3 persons selected are neither Mark nor Ann
then M be the event of selecting M
A be the event of selecting A. Then A^c be the event she is not selected = 1-A
Probability of selecting 4 people = Number of ways of selecting first 3 x number of ways of selecting Mark given that Anna is not selected.
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P(E)= 6C3/8/C4
P(M|A^c) = means that of the remaining 5 persons to choose from, Anna cannot be chosen and so cannot the others b/c we must have Mark.
Let's evaluate P(M|A^c) which you request.
P(M|A^c) = P(M intercept A^c) /P(A^c)
Given the indepdence of the events, this becomes
P(M|A^c) = P(M intercept A^c) /P(A^c) = P(M) P(A^c)/P(A^c) =P(M)
Now we know P(M) is a certainty since we already have the other three people and the probabability of certainty is 1.
That was the solution he earlier provided. I am just giving it the explanation you wanted.
