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A jar contains $m + n$ chips, numbered $1$, $2$, $\ldots$, $n + m$. A set of size $n$ is drawn. If we let $X$ denote the number of chips drawn having numbers that exceed each of the numbers of those remaining, compute the probability mass function.

Suppose we select $n$ chips so the probability will $n/(n+m)$. Can we proceed further in this way or I am doing it incorrectly.

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  • $\begingroup$ Question: What do you mean when you say "let $X$ denote the number of chips drawn having number that exceed each of the numbers remaining? $\endgroup$ – Sentinel135 Feb 22 '17 at 8:04
  • $\begingroup$ suppose there are 9 chips in total, hence X will be greater than 4. That is what i can make out from this statement. $\endgroup$ – Lalit Pathak Feb 22 '17 at 8:06
  • $\begingroup$ Yeah... That's not what I was getting. I was getting that if we pick say $b$ out of the bag. this would mean that all other numbers are less than it (i.e. $b>a$ s.t. $\forall a \in$ not picked). $\endgroup$ – Sentinel135 Feb 22 '17 at 8:10
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This seems to be a draw without replacement.

Suppose $K$ is the largest value remaining in the bag. So all $X=n+m-K$ larger values are drawn and so too are $K-m$ smaller values from the remaining $K-1$ possibilities.

This happens with probability $$P(X=x)=\dfrac{k-1 \choose k-m}{n+m \choose n} = \dfrac{n+m-x-1 \choose n-x}{n+m \choose n}= m \dfrac{(n+m-x-1)! n! }{(n-x)! (n+m)!}$$

giving positive values for $0 \le X \le n$.

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  • $\begingroup$ Taking zoli's example of $m=3$ and $n=2$, so with two balls drawn from a bag of five, this would give $P(X=0)=0.6, P(X=1)=0.3, P(X=2)=0.1$, which looks sensible $\endgroup$ – Henry Feb 22 '17 at 8:43
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I'll try to show my initial probability relationshps. First off, I assume you know what a PMF is or that you can find out what it is. So all we need to figure out is $P(x)$ for all possible $x\in X$.

Using how I interpreted it, if $x=1$ then out of all $n$ picked we had some how picked $(n+m)$. It's not that it's impossible, clearly we can do it, it's like picking the Ace of spades from a 5 card hand. That should give you a hint as to what the probability will be like.

Lets look at the example. there are $(^{52}_5)$ total possible ways of having a 5 card hand. For the ace of spades there is $(^1_1)$ possible ways of getting the ace of spades, and $(^{51}_4)$ ways to pick the rest. Thus for the example the probability of getting the ace of spades in a 5 card hand is $$\frac{(^1_1)*(^{51}_4)}{(^{52}_5)}$$

From the example we can see that $P(1)$ should be very similar. There are $(^{n+m}_n)$ total possible ways of picking $n$ numbers from an $(n+m)$ bag of chips. There are $(^1_1)$ possible ways of getting $n+m$ from the bag, and $(^{n+m-1}_{n-1})$ ways of picking the rest. This implies that for $P(x=1)$ we have, $$P(x=1)=\frac{(^1_1)*(^{n+m-1}_{n-1})}{(^{n+m}_n)}$$

From here you can probably do the rest.

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