Probability with candies

Question

My problem is: A jar containing $1000$ candies, $750$ are red and $250$ are yellow. If you randomly select $10$ candies from the jar, record the amount of red and yellow then replace the candies and repeat this process $50$ times how many times do you think you will get $0$ red candies, $1$ red candy, $2$ red candies,..., and $10$ red candies?

I thought I was doing this right but I seem to have run into an issue that makes me think I am doing it wrong. if I am calculating the amount of times we draw 10 red candies I use this process and a similar on that for other amounts of red candies: $(750!*990!)/(740!*1000!)$ which gets the probability of drawing 10 red candies and $(750!*250!*990!)/(741!*249!*1000!)$ for the chance of getting 9 red candies, and so on. Then I would take that probability and multiply it by 50 to get the number of times I would draw 10 red candies in my 50 trials and so on for the other cases but those numbers are far too small. When I sum all of those percentages that I calculated I got a number far less than $100\%$ which means I cannot use those percentages to find out what number of trials out of 50 would contain 10 red candies. Where am I going wrong? Thank You!

Answer 1

Your formula for all $10$ candies being red is correct. However, the formula you give for exactly $9$ red candies is not correct. What you have written actually calculates the probability that the first $9$ candies are red and the last is yellow - it can equivalently be written as $$\frac{750}{1000}.\frac{749}{999}.\frac{748}{998}.\frac{747}{997}.\frac{746}{996}.\frac{745}{995}.\frac{744}{994}.\frac{743}{993}.\frac{742}{992}.\frac{250}{991}.$$

Since any of the $10$ candies could be the yellow one, you need to multiply this probability by $10$. In general, for $r$ red candies, you will need to multiply by $\binom{10}r=\frac{10!}{r!(10-r)!}$.

Answer 2

For $k\in\{0,1,2,\dots,10\}$ and $i\in\{1,2,\dots 50\}$ let $X_{k,i}$ take value $1$ if at the $i$-th time exactly $k$ red candies are selected.

Then $X_k:=\sum_{i=1}^{50}X_{k,i}$ equals the number of times that exactly $k$ red candies were selected.

With linearity of expectation and symmetry we find:$$\mathsf EX_k=50\mathsf EX_{k,1}=50\mathsf P(X_{k,1}=1)=50\frac{\binom{750}{k}\binom{250}{10-k}}{\binom{1000}{10}}$$

Here $\sum_{k=0}^{10}\binom{750}{k}\binom{250}{10-k}=\binom{1000}{10}$ so that $\sum_{k=0}^{10}\mathsf EX_k=50$ as was to be expected.