Task:
In box are three cards. First card is red on both sides. Second card is black on both sides. Third card is red on one side and black on other side. I random choose one card and I see that color on one side is red. What is the probability that second side is red?
My solution is $P=\frac{1}{2}$ because on other side should be black or red color. Is this correct?
The first card has sides $R_1$ and $R_2$, two red sides. The second has $B_1$ and $B_2$ (two black sides), and the last has $R_3$ and $B_3$.
You see either $R_1$, $R_2$ or $R_3$ and all are equally likely. But $R_1$ has $R_2$ on the other side, $R_2$ has $R_1$ on the other side, and $R_3$ has $B_3$ on the other side. So the other side is red with chance $\frac{2}{3}$.
Let $B$ be the event that you selected the card with different sides, and $R$ that you selected the red card. You are calculating $$P(\text{down red} | \text{up red})=P(B|\text{red up})P(\text{down red}|B)+P(R|\text{red up})P(\text{down red}|R)=\frac{P(B\cap\text{red up})}{P(\text{red up}}\cdot 1+\frac{P(R\cap\text{red up})}{P(\text{red up}}\cdot 0=\frac{P(B)}{P(\text{red up})}=\frac{1/3}{1/2}=2/3$$
You have 3 possibility : you are looking the First card's side 1, you are looking the first card's side 2 or you are looking the third card's side red.
Sot the probability that the second side is red is $\frac 2 3$
