Answer :
(a)let A=event that first k-1 draws will get r-1 red balls
let B=event that last (kth) draw will get rth red ball
P(kth draw is the rth red ball)= P(A AND B) = P(A)xP(B)
(b)P(A)= # of ways to draw r-1 red balls in k-1 trail
# of ways to draw k-1 balls
RBBB *R is one way to achieve P(A AND B)
BBBR *R is another way.
(c)There are n−(r−1) red left, and a total of (m+n)−(k−1) balls left, so the (conditional) probability of drawing a red on the kth trial is n−r+1/m+n−k+1 Thus our required probability is
So, P(A)= nC(r-1) x mC(n-r) x 1/(n+m)C(k-1)
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