Q.3.56 - A First Course in Probability

Question

Suppose that you continually collect coupons and that there are $m$ different types. Suppose also that each time a new coupon is obtained, it is a type $i$ coupon with probability $p_i,i=1,\dots ,m$. Suppose that you have just collected your $n$th coupon. What is the probability that it is a new type?

Hint: Condition on the type of this coupon.

Step-by-Step Solution

Answer

$P=\sum _{i=1}^m{p}_i{\left(1-p_i\right)}^{n-1}$

The wanted event is union of mutually exclusive events the $n-th$ collected coupon is of type $1,2,\dots$or $m$. First, second, third ... collected coupon is of type $i$ are independent events.

Step 1::Given information

Given in the question that that you continually collect coupons and that there are m different types. Suppose also that each time a new coupon is obtained, it is a type i coupon with probability

$p_i,i=1,\dots ,m$

Step 2:Explanation

Events:

$E_{i,j}-i$ th collected coupon is of type $j$

$i\in \mathrm{\mathbb{N}},j\in 1,2,\dots ,m$

Probabilities:

$p_i=P\left(E_{k,i}\right)$ for every $k\in \mathrm{\mathbb{N}}$

Since probabilities are constant, the type of the new coupon is independent of the previous coupons

Calculate $P$- the probability that$n-th$ the collected coupon is the first of its kind.

Step 3: Find the Probability

There are$m$mutually exclusive events that satisfy that $n-th$ collected coupon is the first of its kind - that the $n-th$ coupon is of type $1$, and no coupon of that kind is collected until then, that $n-th$ coupon is no. $2$, and no such coupons are collected until then...

$P=\sum _{i=1}^{m}P\left(E_{1,i}^c{E}_{2,i}^c\dots E_{n-1,i}^c{E}_{n,i}\right)$

$E_{k,i}^c$ is the event that the $k-th$ stamp is not of the $i-th$ kind

Because of the independence:

$P\left(E_{1,i}^c{E}_{2,i}^c\dots E_{n-1,i}^c{E}_{n,i}\right)=P\left(E_{1,i}^c\right)·P\left(E_{2,i}^c\right)·\dots ·P\left(E_{n-1,i}^c\right)·P\left(E_{n,i}\right)$

The formula for the probability of a complement is$P\left(E^c\right)=1-P\left(E\right)$ 

Therefore:

$P\left(E_{1,i}^c\right)·P\left(E_{2,i}^c\right)·\dots ·P\left(E_{n-1,i}^c\right)={\left(1-p_i\right)}^{n-1}·p_i$

We get,

$P=\sum _{i=1}^m{p}_i{\left(1-p_i\right)}^{n-1}$

Step 4: Final answer

$P=\sum _{i=1}^m{p}_i{\left(1-p_i\right)}^{n-1}$

The wanted event is union of mutually exclusive events the $n-th$ collected coupon is of type $1,2,\dots$or $m$.

First, second, third ... collected coupon is of type $i$ are independent events.