Determining color of ball from urn with condition that the two next balls are white

Question

One ball is removed from an urn containing $a$ white and $b$ black balls. To determine the color of the removed ball, two more balls are drawn. What is the probability that a black ball has been removed at the beginning, if both balls drawn afterwards are white?

Answer 1

To simplify the required conditional probability by using notation of colored balls drawn in sequence, and counting number of ways rather than probability (since the denominator will remain the same)

$P(\text{black first} \mid\text{both white afterwards})$

$={P(BWW)\over P (BWW) +P(WWW)}$

$= {n(BWW)\over n(BWW)+ n(WWW)}$

$= \dfrac {b a (a-1)}{b a (a-1) + a (a-1) (a-2)} =\boxed{ \dfrac{b}{b+a-2}}$

Answer 2

Remove the two whites, then you pull a black with probability $\frac{b}{a+b-2}$.

Answer 3

We apply Bayes theorem to calculate the probability:

$$P(\text{black first} \mid\text{both white afterwards})=\dfrac{P(\text{both white afterwards }\mid\text{black first})\cdot P(\text{black first})}{P(\text{both white afterwards})}$$.

Now the individual probabilities. The probability that black will be drawn first is $$ P(\text{black first})=\frac{b}{a+b} $$ because there are a total of $a+b$ balls from which $b$ are black. The probability that both are white afterwards means that there are only $a+b-1$ balls left from the $a+b$ balls: $$ P(\text{both white afterwards} \mid \text{black first})=\frac{a}{a+b-1}\cdot \frac{a-1}{a+b-2}. $$ Finally we calculate $$ P(\text{both white afterwards})=\frac{b}{a+b}\frac{a}{a+b-1}\cdot \frac{a-1}{a+b-2}+\frac{a}{a+b}\frac{a-1}{a+b-1}\cdot \frac{a-2}{a+b-2} $$ Thus we have $$ \dfrac{\frac{a}{a+b-1}\cdot \frac{a-1}{a+b-2}\cdot \frac{b}{a+b}}{\frac{b}{a+b}\frac{a}{a+b-1}\cdot \frac{a-1}{a+b-2}+\frac{a}{a+b}\frac{a-1}{a+b-1}\cdot \frac{a-2}{a+b-2}}= \dfrac{b}{b+a-2}. $$