Probability - Men and women meeting

[Robin04]

Homework Statement

There are N women and M men on a party. They can make friends with the same gender one by one independently with $p$ probability. Men and women cannot make friends. What's the probability that if we choose a man and a woman, the sum of their friends they got to know is $k$.

Relevant Equations

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I'm not sure how to start this. Can you give me a little hint?

 

[scottdave]

Try considering it is just M number of people with probability p. For example say it's 3 people. So a person can have either 1 friend or 2 friends. Once you figure you this situation works, then consider men as a group, pick 1, then women is a separate group, pick 1.

 

[lomidrevo]

Are you familiar with discrete random variables? This particular problem could be solved by applying binomial distribution.

 

[Robin04]

Are you familiar with discrete random variables? This particular problem could be solved by applying binomial distribution.

So the variable that stands for the number of friends of a man or woman follows the binomial distribution. Wikipedia: $\Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$ where $n$ can be M and N, and $p$ is the same $p$ as in our problem. We have now two independent variables. How to get that probability that their sum is a certain value?

 

[PeroK]

So the variable that stands for the number of friends of a man or woman follows the binomial distribution. Wikipedia: $\Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$ where $n$ can be M and N, and $p$ is the same $p$ as in our problem. We have now two independent variables. How to get that probability that their sum is a certain value?

Isn't that what you have calculated?

Hint: can you combine the $M + N$ trials?

 

[Robin04]

Isn't that what you have calculated?

Hint: can you combine the $M + N$ trials?

I found this: https://math.stackexchange.com/questions/1176385/sum-of-two-independent-binomial-variables
According to this the sum of two binomials is another binomial with the sum of their parameters, so $n$ has to be substituted by $M+N$, right?

 

[PeroK]

I found this: https://math.stackexchange.com/questions/1176385/sum-of-two-independent-binomial-variables
According to this the sum of two binomials is another binomial with the sum of their parameters, so $n$ has to be substituted by $M+N$, right?

If you want a total of $k$ successes from two sets of $M$ and $N$ trials, then isn't that the same as $k$ successes from $M + N$ trials? Assuming they are all independent.

The problem is a bit artificial as why would there be the same probability of making a friend in each case? If, instead, each man tossed a coin $M$ times and each woman tossed a coin $N$ times, then the probablity they have $k$ heads between them is clearly the same as if one person had tossed a coin $M + N$ times.

That said, are $M$ and $N$ correct for the number of trials?

 

[lomidrevo]

so n has to be substituted by M+N, right?

Almost there. Maximally how many friends can one make?

 

[Robin04]

Almost there. Maximally how many friends can one make?

N-1 and M-1

 

[lomidrevo]

That looks better

 

[Robin04]

That looks better

So n is N+M-2? But why?

 

[lomidrevo]

Yes, that should be correct. Binomial distribution ca be seen as a sum of $n$ indepedent Bernoulli trials. Each opportunity for a friendship in this example is a Bernoulli trial. So for $n$ you have to use the total number of opportunities to make a friendship (for the chosen woman and man)