What is $\Bbb E(X)$?

Let there be $$100$$ balls in a box out of which $$50$$ are red and $$50$$ are green. Pick $$75$$ balls at random from the box and throw them away. Now pick one ball at random from the remaining balls in the box. Let $$X$$ be the random variable which takes the value $$100$$ when the ball drawn is red in colour and takes the value $$25$$ if the ball drawn is green in colour. Find the expectation $$\Bbb E(X).$$

My attempt $$:$$ Let $$Y$$ denote the number of red balls thrown away. Then the number of green balls thrown away is $$75-Y.$$ Clearly $$Y \geq 25.$$ So \begin{align*} \Bbb P(X=100) & = \sum\limits_{n=25}^{49} \Bbb P(X=100 \mid Y=n)\ \Bbb P(Y=n) \\ & = \sum\limits_{n=25}^{49} \frac {\binom {50-n} {1}} {\binom {25} {1}} \times \frac {\binom {50} {n}} {\binom {100} {n}} \end{align*} Similarly \begin{align*} \Bbb P(X=25) & = \sum\limits_{n=25}^{49} \Bbb P(X=25 \mid (75-Y) = n)\ \Bbb P((75-Y) = n) \\ & = \sum\limits_{n=25}^{49} \Bbb P(X=25 \mid Y = 75-n)\ \Bbb P(Y = 75-n) \\ & = \sum\limits_{n=25}^{49} \frac {\binom {50-n} {1}} {\binom {25} {1}} \times \frac {\binom {50} {75-n}} {\binom {100} {75-n}} \end{align*}

Then the required expectation would be $$100\ \Bbb P(X=100) + 25\ \Bbb P(X=25).$$ But the computation is very tough. Is there any simpler way to approach the problem?

Any help will be highly appreciated. Thank you very much.

• It seems to me that red and green are indistinguishable in this scenario. Whatever the probability of picking a red ball is, it must be equal to the probability of picking a green ball. So P(X=100) = P(X=25) = 1/2 – Ant Jun 13 '20 at 8:51
• But @Ant after throwing $75$ balls away it might not be the case. Obviously the green and red balls in the box are present with different proportions after throwing $75$ balls. Right? Then the red and green balls are not equally likely to be drawn. – Phi beta kappa Jun 13 '20 at 8:55
• I have posted an answer, so we can keep discussing it there. To answer your question, they will be present in a different proportion after being drawn, but that's just a single observation. Their distribution is the same; the distribution of how many red vs green balls end up in the urn after you throw some away is the same. Therefore the distribution of which ball you pick after is also the same. So probabilities are the same. Right? – Ant Jun 13 '20 at 9:02
• It is clear by symmetry that $$\mathbb P(X=25)=\mathbb P(X=100)=\frac12.$$ There is no need for elaborate computation here. – Math1000 Jun 13 '20 at 10:09
• @Math1000 can you show it mathematically without making some vague assertion? I know that you are a genius; but it's better not to think others as genius as you. – Phi beta kappa Jun 13 '20 at 10:22

It seems to me that red and green are indistinguishable in this scenario. Whatever the probability of picking a red ball is, it must be equal to the probability of picking a green ball. So $$P(X=100) = P(X=25) = 1/2$$

Which implies

$$E(X) = 100\cdot P(X=100) + 25 \cdot P(X=25) = 62.5$$

A simple script confirms that the expectation is indeed $$62.5$$. You can also double check by computing the probability. Using your terminology,

$$P(Y=n) = \frac{{75 \choose n} \cdot {25 \choose 50-n}} {100\choose 50}$$

So that

$$P(X=100) = \sum_{n=25}^{50} \frac{50-n}{25} \cdot \frac{{75 \choose n} \cdot {25 \choose 50-n}} {100\choose 50}$$

And you can check numerically that this is equal to $$1/2$$

• How have you computed $\Bbb P(Y=n)$? – Phi beta kappa Jun 13 '20 at 9:43
• The probability of removing n red balls can be computed as follows: Line up all the 100 balls in a row. Count how many of the permutations end up with n balls in the first 75 spots. Divide by total number of permutations, and you're done. To compute the first number, you have 75 balls of which n are red. So the permutations are 75!/n!/(75-n)!, i.e. 75 choose n. Then you have to multiply with the number of permutations of the remaining 25 balls; it's the same, except that you now have 50-n red balls. The second number is the total number of permutations, therefore 100 choose 50. – Ant Jun 13 '20 at 9:48
• I don't think so. If according to your assumption the balls are indistinguishable then we have just $26$ ways to throw $75$ balls from the urn and each way is equally likely. If the balls are indistinguishable then it doesn't matter in which order you throw them away or which red/green balls are thrown away. In this case which really matters that how many red/green balls you throw away out of $75$ balls. Right? If I present them as ordered pairs as (number of red balls thrown away, number of green balls thrown away) then there are $26$ such pairs $(25,50),(26,50), \cdots, (50,25).$ – Phi beta kappa Jun 13 '20 at 10:15
• Each pair is equally likely to occur. So for each $25 \leq n \leq 50$ we have $$\Bbb P(Y=n) = \frac {1} {26}.$$ – Phi beta kappa Jun 13 '20 at 10:17
• No. I know it for the case of distinguishable but not for the case of indistinguishable. – Phi beta kappa Jun 13 '20 at 10:28