I was solving a probability exercise, there is 2 jars $U_1$ and $U_2$ each containing $10$ balls.
In $U_1$ there is $6$ red balls and $4$ black balls. In $U_2$ there is $5$ red balls and $5$ black balls.
A player rolls a fair dice of 6 facing, if he get the numbers 1 or 2, he chooses 2 balls simultaneously from jar $U_1$, if he gets 3,4,5, or 6 he chooses 2 balls from jar $U_2$ randomly and with replacement.
Consider the following events:
$U_1:$The chosen balls are from jar $U_1$
$R:$ The chosen balls are red
A part of the question is to calculate $P(R \cap U_1)$ (i.e. the probability that the chosen balls are red and from jar $U_1$). I worked out the answer using the rule $$P(R\cap U_1)=P(R/U_1)×P(U_1)=\frac{1}{9}$$
But I was thinking about another way to calculate it without using this rule, I thought about $$P(R\cap U_1)=\frac{\frac{1}{3}C_6^2}{\frac{1}{3}C_{10}^2+\frac{2}{3}×10^2}$$
Because we want 2 red balls from jar $U_1$ and that is $\frac{1}{3}C_6^2$ divided by the total number of ways that we can choose 2 balls from $U_1$ or $U_2$ and that is $\frac{1}{3}C_{10}^2$ (for $U_1$) and $\frac{2}{3}×10^2$ (for $U_2$).
But it seems to be wrong, as this gave a different answer than the first one, and ofcourse the first answer is right, but I can't see what I missed? And how can this be solved without using the first mentioned rule?
Thank you for you help...