Intersection of events in Coupon Collector Problem $ P(A_i \cap. A_j) $

Question

I am studying a variation of the coupon collector problem where there are N coupons, and each coupon has a probability $p_i$ of appearing with $i = 1,2,3 ... N$. I have to find $P(T=n)$ where T is a random variable that defines the number of tries until we have a complete collection of coupons (one of each of the N). To do so, we find $P(T>n)$ instead.

While I understand the idea of defining $A_i = \{\text{coupon i did not come up in the n tries}\}$, I don't understand why $P(A_i \cap A_j) = (1 - (p_i+p_j))^n$. Shouldn't it be $P(A_i \cap A_j)= ((1-p_i)(1-p_j))^n$. This because each extraction is independent from the other, as is each coupon from the other.

In summary, why is $P(A_i \cap A_j) = (1 - (p_i+p_j))^n$? As in, what is the underlying logic or the way to derive that formula (I understand the raising to the n part given that it happens n times, I don't get the $(1-(p_i+p_j))$. ) Thanks!

(BTW, this is in Ross's A first course in probability, Theoretical exercise 4.1)

Answer 1

Assuming $i\not=j$

• $p_i$ is the probability coupon $i$ turns up on a particular try, event $A_i$
• $p_j$ is the probability coupon $j$ turns up on a particular try, event $A_j$
• These are mutually exclusive events since $P(A_i \cap A_j)=0$ and $P(A_i \cup A_j)=P(A_i)+P(A_j)$, not independent events which would have $P(A_i \cap A_j)=P(A_i)\,P(A_j)$ and $P(A_i \cup A_j)=P(A_i)+P(A_j)-P(A_i)\,P(A_j)$
• So $p_i+p_j$ is the probability one of coupon $i$ and coupon $j$ turns up on a particular try
• And $1-(p_i+p_j)$ is the probability neither coupon $i$ nor coupon $j$ turns up on a particular try
• Successive tries are independent, so $(1-(p_i+p_j))^n$ is the probability neither coupon $i$ nor coupon $j$ turns up in any of $n$ tries

Answer 2

While I understand the idea of defining $A_i = \{\text{coupon i did not come up in the n tries}\}$, I don't understand why $P(A_i \cap A_j) = (1 - (p_i+p_j))^n$. Shouldn't it be $P(A_i \cap A_j)= ((1-p_i)(1-p_j))^n$. This because each extraction is independent from the other, as is each coupon from the other.

First of all,

$$P(A_i \cap A_j)= ((1-p_i)(1-p_j))^n$$

is wrong, because on a given turn, getting Coupon-i and getting Coupon-j are not independent events. That is, on a given turn, when you don't get Coupon-i, it is slightly more likely than normal that you do get Coupon-j.

Then,

$$(1 - (p_i+p_j))$$

represents the chance of not getting either Coupon-i or Coupon-j on 1 specific turn. Therefore,

$$(1 - (p_i+p_j))^n$$

represents the chance of getting not getting either Coupon-i or Coupon-j on $(n)$ consecutive turns.