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 $$P(X=100) = P(X=25) = 1/2$$
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 $1/2$