Tom K. answered 10/15/20
Knowledgeable and Friendly Math and Statistics Tutor
There are many ways of reasoning this, but the probability is m/n
Three ways to show. Among them,
One: it's obvious
Two: Let the first person be P1, the second person P2, ..., the nth person Pn
E(number selected) = E(P1 + P2 + ...+ Pn) = E(P1) + E(P2) + ... + E(Pn)
E(number selected) is m, and E(P1) = E(P2) = E(P3) ... = E(Pn), so n E(P1) = m, or E(P1) = m/n, and there is the same probability for each person, including you.
Three: There are C(n,m) = n!/(m!n-m!) ways to select the committee
If you are selected, m - 1 of the remaining n-1 must be selected, so the number of ways of selecting you is
C(n-1,m-1) = n-1!/((m-1!)(n-1 - (m-1))!) = n-1!/((m-1!)(n-m)!), so
P(you are selected) = n-1!/((m-1!)(n-m)!)/( n!/(m!n-m!)) =
n-1! m!n-m!/(m-1!n-m!n!) =
n-1!m!/(m-1!n!) =
n-1!m*m-1!/(m-1!n*n-1!) =
m/n