# Probability: Combinatorics and discrete random variables I've managed part (i) fine but have no idea how to approach part (ii). I tried using Baye's theorem in order to calculate the conditional probability that the red team has size k given that it contains A but ended up with a probability greater than 1...

• First, Why don't you tell us what you got for (i) as part of your Question; I believe you, but maybe some insurance against down-votes and getting your question put on Hold. Second, it's Bayes' Theorem (after Rev. Thomas Bayes). Third, the formulas at the end are a major clue. Last, it usually helps to consider a special case to get started; see my outline below, which is not really an Answer, but may help you find one. – BruceET Jun 17 '15 at 20:42

Outline for the special case $n = 10.$

Then the red team can be of sizes $x = 1, \dots, (n-1) = 9,$ and each of the nine sizes is equally likely. Symmetrically, for the other team.

For (ii), Player A is more likely to be chosen if he is in a larger team. Argue that the probabilities for Player A's team to be of sizes $k = 1, \dots , 9$ are $k/45,$ respectively, where the denominator comes from one of the formulas at the end of your question.

Then use the other summation formula to get 19/3 as the answer to (iii), for $n = 10.$

If you can make a rigorous argument for $n = 10,$ go on to the general problem. If not, maybe consider case $n = 4,$ for which you should be able to make a list of the possibilities.

Finally, generalize to any $n \ge 2.$

Addendum: Below is a simulation in R for a million performances of this experiment with 10 players. It illustrates (mainly with 2 or 3 place accuracy) some some of the results you should obtain using probability rules.

 m = 10^6;  n = 10;  size.a = size.red = numeric(m)
for(i in 1:m) {
x = sample(1:(n-1), 1);  a = sample(1:n, 1);  red = 1:x
size.a[i] = size.red[i] = length(red)  # if A on red team
if (a > x) size.a[i] = n - x }         # if not

round(table(size.a)/m, 3);  mean(size.a) # info on A's team
## size.a
##     1     2     3     4     5     6     7     8     9
## 0.022 0.045 0.067 0.089 0.111 0.134 0.156 0.178 0.199
##  6.33151

round(table(size.red)/m, 3);  mean(size.red)  # info on red tesm
## size.red
##     1     2     3     4     5     6     7     8     9
## 0.111 0.111 0.111 0.111 0.111 0.112 0.111 0.112 0.111
##  5.002948


Break it down. For Player $A$ to even have a chance of being on a team with $k$ players (for $k<n$), you need $X=k$ or $X=n-k$. These two happen with probability $1/(n-1)$. Start with $X=k$. We need player $A$ to land in the first team, which happens with probability $\binom{n-1}{k-1}/\binom{n}{k}=\frac{k}{n}$: we set aside Player A and pick the rest of the team. The other option is when $X=n-k$ which works out similarly. One thing to be careful about is if $n=2m$ and $k=m$ in which case Player $A$ has probability 1 of being on such a team.

• Just referred you to this question, related to one of your own from a day or two ago. Glad you discovered it on your own. – BruceET Jun 17 '15 at 20:58
• @BruceTrumbo: A very strange coincidence indeed! The funny thing is I posted this before seeing your comment on my other question. – Alex R. Jun 17 '15 at 22:27