Probability of drawing $3$ red marbles from $5$ red and $10$ blue marbles What is the probability that I will draw only $3$ red marbles from the mix of $5$ red and $10$ blue marbles if I draw $5$ times?
Is it just p(at least $3$ reds)$\cdot$ p($2$ blues given $3$ reds) 
Where probability of at least $3$ reds = $\Large\frac{5}{15} \cdot \frac{4}{14}\cdot \frac{3}{13} $
And probability $2$ blues given $3$ reds = $\Large\frac{10}{12}\cdot \frac{9}{11}$ 
 A: Some hints:


*

*Choose the three red marbles out of five.  How many ways can you do this?

*Choose the two blue ones out of ten.  How many ways can you do this?

*How many ways can you choose five marbles from fifteen (regardless of how many reds and blues)?


Answer below:

 $$\frac{{5 \choose 3}{10 \choose 2}}{{15 \choose 5}}$$

A: The things you calculated were not what you claimed you were calculating.   Yet somehow you were close to the correct answer. 
(1) "only three" means "exactly three" not "at least three".
(2) You gave the probability of drawing exactly three red marbles among three marbles drawn, not as you claimed, the probability of drawing at least three red marbles among five.
(3) However you also gave the probability of drawing two blue marbles among an additional two drawn given 3 red were previously removed.  You can actually combine these answers with the count of ways to arrange the five marbles (two blue, three red) to get the correct answer.
The probability of drawing some arrangement of 3 red and 2 blue marbles among 5 marbles drawn is: the count of permutations times the probability of drawing any particular arrangement of 3 red and 2 blue.

 $$\dbinom{5}{3}\left(\dfrac 5 {15}\dfrac 4{14}\dfrac 3{13} \right)\left(\dfrac {10}{12}\dfrac 9{11}\right)$$

