Selection Probability

Question

One urn contains one blue ball (labeled $B_1$) and three red balls (labeled $R_1$, $R_2$, and $R_3$). A second urn contains two red balls ($R_4$ and $R_5$) and two blue balls ($B_2$ and $B_3$). An experiment is performed in which one of the two urns is chosen at random and then two balls are randomly chosen from it, one after the other without replacement.

What is the probability that two red balls are chosen?

I found the total number of possibilities to be $24$, so I thought the answer to this would $\frac {2}{24} = \frac {1}{12}$ but it's apparently not correct so I'm at a loss.

Answer 1

Ok, let's start by counting.

If we choose the first urn, what is the probability that we gen 2 red balls? We have in there 1 blue and 3 red balls, so the probability to get 2 reds is: $$p_1=\frac{3}{4}\cdot \frac{2}{3} $$

For the secong urn we get: $$p_2=\frac{2}{4}\cdot\frac{1}{3} $$

And in the end just sums it all up by caring about probability of choosing each urn. For both of them we have $\frac{1}{2}.$ So the answer is: $$p=\frac{1}{2}p_1 +\frac{1}{2}p_2=\frac{1}{3}$$

Answer 2

Let Urn 1= $B_1,R_1,R_2,R_3$ and Urn 2= $B_2,B_3,R_4,R_5$
$P(A)$=Probability that 2 balls are chosen Red
$P(U_1)$=Probability that Urn 1 Is taken (0.5)
$P(U_2)$=Probability that Urn 2 Is taken (0.5)

Your answer: $P(A|U_1)+P(A|U_2)$ $$\frac{^3C_2*0.5}{^4C_2}+\frac{^2C_2*0.5}{^4C_2}$$

This should give your answer