A box contains $10$ blue marbles and $2$ red marbles. $5$ marbles are drawn at random with replacement. What is the chance that both the red marbles will be drawn?
Is my solution correct?
$$ \frac{{2 \choose 2} \cdot {10 \choose 3}}{{12 \choose 5}}\ $$
${2 \choose 2}$ for the number of ways to arrange $2$ red marbles, ${10 \choose 3}$ for number of ways to choose $3$ blue marbles over the total number of ways to choose $5$ marbles.
 A: Let $X_1$ denote the number of times we draw the first red marble, $X_2$ the number of times we draw the second red marble, and $X_3$ be the number of times we draw a blue marble. We wish to find $$P(X_1 \geq 1, X_2 \geq 1)$$
Notice that since we're dealing with successes of more than one category, and we are replacing the marbles, that this is a multinomial distribution:
$$P(X_1=x_1, X_2=x_2, X_3=x_3)=\frac{n!}{x_1!\cdot x_2!\cdot x_3!}\cdot p_1^{x_1}\cdot p_2^{x_2}\cdot p_3^{x_3}$$
where $$p_1=p_2=\frac{1}{12}, p_3=\frac{5}{6}$$
You must sum for all possible values of $x_1,x_2,x_3$ which satisfy the criteria, and you should find there are $10$ viable combinations. 
I would suggest making these calculations in excel. As a check of your work, I will give you that an R simulation gives a probability of roughly $0.107$ which agrees with the analytical result I obtained.
coin=c(rep("B",10),"RedOne","RedTwo")
u=replicate(10^6,sample(coin,5,repl=T))
blue = colSums(u=="B")
redone =colSums(u=="RedOne")
redtwo=colSums(u=="RedTwo")
mean(redone >= 1 & redtwo >= 1) 

[1] 0.107308

