# Probability problem: Correct me if I am wrong.

There are two urns $U_1$ and $U_2$. $U_1$ contains four white and four black balls, and $U_2$ is empty. Four balls are drawn at random from $U_1$ and transferred to $U_2$. Then a ball is drawn at random from $U_2$. Find the probability that the ball drawn from $U_2$ is white.

I did Let $U_w$ be the event that the ball drawn is white. $k$ be the number of white balls among the four transferred balls
$$P(U_{w})=(1/4)P(k=1)+(2/4)P(k=2)+(3/4)P(k=3)+(4/4)P(k=4)$$ the value on the left hand side of every $P(k=i)$ is the probability of white ball out of 4 balls in $U_2$ all the $P(k=i)$ I find out by using hypergeometric distribution formula and I get the solution $=\frac{1}{2}$ so I want to ask is it the correct approach, how can i do it much more easily and faster. thanks

• Each ball is equally likely to be the one eventually drawn. – lulu Feb 3 '16 at 19:37
• seriously is it that simple did you mean the ball eventually drawn is white has probability $\frac{4}{8}$? just this. please explain your comment a bit – Onix Feb 3 '16 at 19:41
• Intuitively: Replace each white ball by a black ball and each black ball by a white ball. In the new problem, the probability you got should be the probability of getting a black ball ... ;) – N. S. Feb 3 '16 at 19:41
• @Abomm Your solution is good and it's really the right way to do it. You partitioned the probability space by $k$ and then you write $P(U_w)$ in terms of that partition, using conditional probabilities. That's good. Of course you could just intuit the answer because it's fairly obvious intuitively. But it won't usually be intuitive and often intuition is wrong, so it's best to do it the way you did which is precise and rigorous. – Gregory Grant Feb 3 '16 at 20:36

Easier to work from symmetry. Nothing in the selection process favors one ball over another, so each ball has the same probability ($\frac 18$) of being the one drawn. As half the balls are white the probability that the drawn ball is white is $\frac 12$.
• Symmetry is that each ball is treated equally. Number them $\{1,\dots 8\}$. How could it be that #$3$ was more probable than #$5$? Whatever paths lead to the drawing of #$3$ lead to #$5$ if you just swap the two numbers. If you doubt it, follow your method. Compute the probability that #$1$ is chosen. – lulu Feb 3 '16 at 19:48