Probability with candies 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!

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 $\binom{10}r=\frac{10!}{r!(10-r)!}$.
