Results 1 to 2 of 2

Thread: Collecting coupons

  1. #1
    Member
    Joined
    Oct 2006
    Posts
    206
    Thanks
    1

    Collecting coupons

    Suppose that there are $\displaystyle N$ 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 $\displaystyle N$ types. One random variable of interest is $\displaystyle T$, 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 $\displaystyle P\{T=n\}$ directly, let us start by considering the probability that $\displaystyle T$ is greater than $\displaystyle n$. To do so, fix $\displaystyle n$ and define the events $\displaystyle A_{1}, A_{2}, ..., A_{N}$ as follows: $\displaystyle A_{j}$ is the event that no type $\displaystyle j$ coupon is contained among the first $\displaystyle n$, $\displaystyle j=1, ..., N$.
    Hence, $\displaystyle P\{T>n\}=P\left(\bigcup^{N}_{j=1}A_{j}\right)$

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

    Appreciate those who help!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by acc100jt View Post
    Suppose that there are $\displaystyle N$ 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 $\displaystyle N$ types. One random variable of interest is $\displaystyle T$, 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 $\displaystyle P\{T=n\}$ directly, let us start by considering the probability that $\displaystyle T$ is greater than $\displaystyle n$. To do so, fix $\displaystyle n$ and define the events $\displaystyle A_{1}, A_{2}, ..., A_{N}$ as follows: $\displaystyle A_{j}$ is the event that no type $\displaystyle j$ coupon is contained among the first $\displaystyle n$, $\displaystyle j=1, ..., N$.
    Hence, $\displaystyle P\{T>n\}=P\left(\bigcup^{N}_{j=1}A_{j}\right)$

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

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

    About $\displaystyle P(T=n)$, 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 $\displaystyle P(T>n)$ and then deduce the first one by $\displaystyle P(T=n)=P(T>n-1)-P(T>n)$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Expected Value of Collecting a set
    Posted in the Statistics Forum
    Replies: 1
    Last Post: Nov 24th 2011, 04:04 PM
  2. expected number of coupons
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: Feb 18th 2011, 02:11 AM
  3. Stat Problem - Coupons
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: May 18th 2010, 07:02 AM
  4. collecting cards
    Posted in the Statistics Forum
    Replies: 2
    Last Post: Nov 13th 2009, 03:40 AM
  5. coupons
    Posted in the Business Math Forum
    Replies: 2
    Last Post: Jun 27th 2009, 01:56 PM

Search Tags


/mathhelpforum @mathhelpforum