Finding the joint probability of dependent events

Question

I am trying to probabilistically model a problem, a simple version of which is as below:

I have 100 participants for a lucky draw contest. Let us name them $a_{1}, a_{2},...a_{100}.$

Suppose that only the 10 participants namely $a_{1}, a_{2},...a_{10}$ are the possible winners and others are just participating for fun.

The lucky draw is as follows:

I will select each of the 100 participants independently with probability $p$ and if, in the selected set of people, there is only 1 participant from the set $\mathcal{D}=\{a_{1}, a_{2},...a_{10}\}$,then that participant is a winner.

I will repeat this procedure (which I call test) $T$ times. and I call my experiment to be a success if $a_{1}, a_{2}$ and $a_{3}$ are the winners after the $T$ tests.

Let,

$L_{i}=\# \text { tests containing } i \text { and no other element of } \mathcal{D}, \text { for } i=a_{1}, \ldots, a_{10}$

$\mathbb{P}(\text { success })=\mathbb{P}\left(L_{1} \neq 0,L_{2} \neq 0, L_{3} \neq 0\right)$

How can I compute this probability of success? The problem I am facing is since $L_{1},L_{2},L_{3}$ are not independent, I am not able to simplify the expression further?

Answer 1

The probability that $a_1$ is "a winner" in a particular test if selected and the other nine are not with probability $p(1-p)^9$.

The same is true of $a_2$ and $a_3$ and these are mutually exclusive events for that particular test, so we can add their probabilities.

So over all $T$ tests:

• $P(L_1=0) = P(L_2=0) = P(L_3=0) = \left(1- p(1-p)^9\right)^T$
• $P(L_1=0, L_2=0) = P(L_2=0,L_3=0) = P(L_3=0,L_1=0) = \left(1- 2p(1-p)^9\right)^T$
• $P(L_1=0, L_2=0, L_3=0)=\left(1- 3p(1-p)^9\right)^T$

So by inclusion-exclusion $P(L_1=0 \text{ or } L_2=0 \text{ or } L_3=0) = 3\left(1- p(1-p)^9\right)^T - 3 \left(1- 2p(1-p)^9\right)^T + \left(1- 3p(1-p)^9\right)^T$

and thus $P(L_1\not =0, L_2\not =0, L_3\not =0) = 1- 3\left(1- p(1-p)^9\right)^T + 3 \left(1- 2p(1-p)^9\right)^T - \left(1- 3p(1-p)^9\right)^T$

This is an increasing function of $T$ but not of $p$. For example

• if $p=\frac1{100}$ and $T=100$ then this is about $0.2142$ as none of them are selected very often
• if $p=\frac1{10}$ and $T=100$ then this is about $0.9432$
• if $p=\frac1{2}$ and $T=100$ then this is about $0.00078$ as too many from $\mathcal D$ are selected too often