What is the probability that it is the only white ball There are $5$ balls in a bag. It is equally likely that $0,1,2,3,4,5$ balls are white i.e., if $X$ is random variable denoting number of white balls it is given that
$$P(X=0)=P(X=1)=\cdots=P(X=5)=\frac{1}{6}$$
Now a ball is drawn at random and is found to be white. What is the Probability that it is the  only white ball in the bag.
My try:
i actually thought the required probability is $P\left(X=1|X \ge 1 \right)$ which is $\frac{\frac{1}{6}}{\frac{5}{6}}=\frac{1}{5}$
But i am not sure. 
 A: This is close, but the event that you are conditioning on is that the ball drawn was white not the condition that there is at least one white ball.
Let $Y$ be the event that the ball drawn was white.  We are tasked with calculating $Pr(X=1\mid Y)$ which is a different number than $Pr(X=1\mid X\geq 1)$.
$Pr(X=1\mid Y)=\frac{Pr((X=1)\cap Y)}{Pr(Y)}$
$=\frac{Pr(X=1)Pr(Y\mid X=1)}{Pr(Y\cap (X=0))+Pr(Y\cap (X=1))+Pr(Y\cap (X=2))+Pr(Y\cap (X=3))+Pr(Y\cap (X=4))+Pr(Y\cap (X=5))}=\dots$
A: If you wanted to visualize it, you could draw a six by five chart (# of white balls (0-5) by # of any ball picked (1-5). If the column has 1 on the top, indicating that the bag has 1 white ball, then put a W in one of the boxes in that column. You are essentially creating the six possibilities for bags containing white balls (1 W ball, 2 W balls, etc.). Since you chose 1 white ball, you only need to look at the squares that contain a W, where a white ball is chosen. There is only one case in which you choose a white ball and it is the only one in its bag. There are a total of 15 cases in which you pick a white ball. I would think the answer would be $\frac{1}{15}$ but I could be wrong.
A: I agree with your thought for the required probability, but I think you may have made an error in the numerator of your calculation. Recall that $P(A\vert B) = \frac{P(A \cap B)}{P(B)}$. What is the probability that $X=1$ and $X \geq 1$ simultaneously?
