Question

There are N distinct types of coupons, and each time one is obtained it will, independently of past choices, be of type i with probability Pi, i = 1, ... , N. Let T denote the number one need select to obtain at least one of each type. Compute P{T = n}.

Short answer

In the given information the answer is$P(T=n)=P(T>n-1)-P(T>n)$

Step-by-step solution

Step 1:Given Information

Consider event$A_i$,which states that we havent picked a coupon of type i in our sample i=1.....N so $T>n$is equal to the information that is satisfied some of the event $A_i$.ie,

$P(T>n)=P\left({\cup }_{i=1}^N{A}_i\right)$

Calculation

$P\left({\cup }_{i=1}^N{A}_i\right)=\sum _{i_1}P\left(A_{i_1}\right)-\sum _{i_1<i_2}P\left(A_{i_1},A_{i_2}\right)+\sum _{i_1<i_2<i_3}P\left(A_{i_1},A_{i_2},A_{i_3}\right)$

$P\left(A_{i_1}\right)={\left(1-p_{i_1}\right)}^n$

$P\left(A_{i_1},A_{i_2}\right)={\left(1-\left(p_{i_1}+p_{i_2}\right)\right)}^n$

$P\left(A_{i_1},A_{i_2},\dots ,A_{i_j}\right)={\left(1-\left(p_{i_1}+p_{i_2}+\cdots +p_{i_j}\right)\right)}^n$

The required probability T=n can be obtained as$P(T=n)=P(T>n-1)-P(T>n)$

Step 3:Final Answer

The final answer is$P(T=n)=P(T>n-1)-P(T>n)$