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$n$ vehicles are stopped at random, the probability that a driver who is stopped is a beginner is $p$ while the probability that a driver who is stopped is a professional is $q$. There are drivers that are neither beginners nor professionals. $X$ is the random variable representing the number of beginners stopped while $Y$ is the number of professional drivers stopped.

  • If $X_k$ is the random variable of the number of beginners stopped knowing that exactly $k$ professionals were also stopped, what is is the conditional probability $\mathrm {Pr}(X=i|Y=k)?$
  • What is the probability $\mathrm {Pr}(X=i, Y=j)$?
  • If $Y_k$ is the variable defined by $\cfrac Y{X+Y}$ with $X+Y = k$, what is the probability distribution and expected value of $Y_k$?
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    @Robert: That was also my first reaction, but if you Google "conditional random variable", you'll find that some authors introduce such a construction. Not that it seems like a good idea...2012-11-05

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If I understand correctly what you mean by a conditional random variable, your probability $\Pr(X_k=i)$ could equivalently be expressed as $\Pr(X=i\mid Y=k)$. If $k$ professionals were stopped, $n-k$ further people were stopped, and each has an independent probability $p/(1-q)$ of being a beginner and $(1-p-q)/(1-q)=1-p/(1-q)$ of being neither a beginner nor a professional. Thus we have a binomial distribution,

$ \begin{align} \Pr(X=i\mid Y=k) &= \binom{n-k}i\left(\frac p{1-q}\right)^i\left(\frac{1-p-q}{1-q}\right)^{n-k-i} \\ &= (1-q)^{-(n-k)}\binom{n-k}ip^i(1-p-q)^{n-k-i}\;. \end{align} $

For the second question, this is just the overall multinomial distribution,

$ \Pr(X=i,Y=j)=\binom n{i,j,n-i-j}p^iq^j(1-p-q)^{n-i-j}\;. $

The third question is again less than fully clear to me but from analogy with the first question, it appears that the intention is to condition on $X+Y=k$. As for the first question, this leads to a binomial distribution, this time with probabilities $p/(p+q)$ and $q/(p+q)$:

$ \begin{align} \Pr(Y=i\mid X+Y=k) &= \binom ki\left(\frac q{p+q}\right)^i\left(\frac p{p+q}\right)^{k-i} \\ &= (p+q)^{-k}\binom kiq^ip^{k-i} \;, \end{align} $

and the expected value of the proportion $Y/(X+Y)$ is simply the probability $q/(p+q)$, independent of $k$, so this is also the expected value of $Y/(X+Y)$ without conditioning.

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    you just gave me one more point in my exam. Thank you!2012-11-05