Question

There are n types of coupons. We collect coupons one-by-one. Each coupon collected is of type i with probability pi P , and independent of other coupons. Assume that n i=1 pi = 1. Suppose in total k coupons are collected. Define Ai to be the event that there is at least one type i coupon among those collected for i = 1, 2, · · · , n.

(a) Compute P(Ai)

(b) For any i 6= j, find P(Ai ∪ Aj )

(c) Compute P(Ai |Aj ), hint: use formula about P(A ∪ B) = . . .

Answer 1

(a) Since A_i is the event that there is atleast one type i coupon among those collected, hence A_i^c is the event that there is no
type i coupon among those collected.
The probability that the first coupon collected is not of type i is 1-p_i.
The probability that the second coupon collected is not of type i is 1-p_i.
......................................
The probability that the k^th coupon collected is not of type i is 1-p_i.

Since the event of collecting a coupon at the j^th (j=1,2,....k) time is independent of the other collections, hence
P(A_i^c) = (1-p_i).(1-p_i)....(1-p_i) = (1-p_i)^k

Thus, P(A_i) = 1 - P(A_i^c) = 1 - (1-p_i)^k


(b) For any i not equal j, A_i A_j is the event that there is atleast one type i coupon or atleast one type j coupon among those
collected. Thus, (A_i A_j)^c = A_i^c A_j^c is the event that there is neither a type i coupon nor a type j coupon among those collected.

The probability that the coupon collected at the t^th(t=1,2,...k) time is either of type i or of type j is p_i+p_j (mutually exclusive
events).
Thus, the probability that the coupon collected at the t^th(t=1,2,..k) time is neither of type i nor of type j is 1-p_i-p_j.
Thus,
  The probability that the first coupon collected is neither of type i nor of type j is 1-p_i-p_j.
  The probability that the second coupon collected is neither of type i nor of type j is 1-p_i-p_j.
..................................................................................................
  The probability that the k^th coupon collected is neither of type i nor of type j is 1-p_i-p_j.

Since the event of collecting a coupon at the j^th (j=1,2,....k) time is independent of the other collections, hence,
P((A_i A_j)^c) = (1-p_i-p_j).(1-p_i-p_j)....(1-p_i-p_j) = (1-p_i-p_j)^k.

Thus, P(A_i A_j) = 1 - P((A_i A_j)^c) = 1 - (1-p_i-p_j)^k.
  


(c) For i not equal j, P(A_i A_j) = P(A_i)+P(A_j)-P(A_i A_j) = 1- (1-p_i)^k + 1 - (1-p_j)^k - 1 + (1-p_i-p_j)^k = 1 - {(1-p_i)^k+(1-p_j)^k-(1-p_i-p_j)^k}
Thus,
P(A_i | A_j) = P(A_i A_j) / P(A_j) = ( 1 - {(1-p_i)^k+(1-p_j)^k-(1-p_i-p_j)^k} ) / ( 1 - (1-p_j)^k )

Hence, P(A_i | A_j) = ( 1 - {(1-p_i)^k+(1-p_j)^k-(1-p_i-p_j)^k} ) / ( 1 - (1-p_j)^k ).