Math Help Forum - View Single Post

Laurent · #2 July 3rd, 2009, 06:50 AM

Quote:

Originally Posted by acc100jt

Suppose that there are $N$ distinct tyoes of coupons and each time one obtains a coupon it is, independent of prior selections, equally likely to be any one of the $N$ types. One random variable of interest is $T$ , the number of coupons that needs to be collected until one obtains a complete set of at least one of each type. Rather than derive $P\{T=n\}$ directly, let us start by considering the probability that $T$ is greater than $n$ . To do so, fix $n$ and define the events $A_{1}, A_{2}, ..., A_{N}$ as follows: $A_{j}$ is the event that no type $j$ coupon is contained among the first $n$ , $j=1, ..., N$ .
Hence, $P\{T>n\}=P\left(\bigcup^{N}_{j=1}A_{j}\right)$

I coulnd't understand the last equality, and why can't we derive $P\{T=n\}$ directly?

Appreciate those who help!!

The equality is just a translation from words to maths of a simple statement, namely "The event $\{T>n\}$ means that among the first n coupons there is a missing type $j\in\{1,\ldots,N\}$ of coupons". This should be clear (otherwise, give it a second thought ; remember $T$ is the first time we have all types of coupons).
Then $\bigcup_{j=1}^N A_j$ is the event "There exists $j\in\{1,\ldots,N\}$ such that $A_j$ happens", and here we are since $A_j$ is defined as: "the type $j$ is missing among the $n$ first coupons".

About $P(T=n)$ , if you can derive it directly, that's just fine! It is probably easier however (here and in a large variety of situations) to find $P(T>n)$ and then deduce the first one by $P(T=n)=P(T>n-1)-P(T>n)$ .