91Ó°ÊÓ

There are $k$types of coupons. Independently of the types of previously collected coupons, each new coupon collected is of type$i$with probability $pi$,$\underset{i=1}{\overset{k}{∑}}{p}_i=1$

Given:

$k$types of coupon.

Coupons are independently selected

Coupon is of type $i$with probability $p_i$

$\underset{i=1}{\overset{k}{∑}}{p}_i=1$

The number of successes among a fixed number of independent trials with a constant probability of success follows a binomial distribution.

Definition binomial probability:

$P(X=k)=\left(\begin{array}{l}n\\ k\end{array}\right)·p^k·(1-p{)}^{n-k}=\frac{n!}{k!(n-k)!}·p^k·(1-p{)}^{n-k}$

Let us evaluate the definition of binomial probability at $n=n,p=p_i$and $k=0$:

$P($no coupons of type $i)=P(X=0)$$=\left(\begin{array}{l}n\\ 0\end{array}\right)·p_i^0·{\left(1-p_i\right)}^{n-0}$

$=1·1·{\left(1-p_i\right)}^n$

$={\left(1-p_i\right)}^n$

Use the Complement rule:

$P\left(A^c\right)=P($not $A)=1-P(A)$

$P($At least one coupon of type $i)=1-P($no coupons of type$i)$

$=1-{\left(1-p_i\right)}^n$

The expected value (or mean) is the sum of the product of each possibility $x$(number of types of coupons) with its probability $P\left(x\right)$.

$E($Number of types $)=∑xP(x)$

$=\underset{i=1}{\overset{k}{∑}}1×\left(1-{\left(1-p_i\right)}^n\right)$

$=\underset{i=1}{\overset{k}{∑}}\left(1-{\left(1-p_i\right)}^n\right)$

$=k-\underset{i=1}{\overset{k}{∑}}{\left(1-p_i\right)}^n$

$E($Number of types$)=$$k-\underset{i=1}{\overset{k}{∑}}{\left(1-p_i\right)}^n$