probability for ball replacement

Question

Three boxes of the same appearance have the following proportion of balls Box I : 2 black 1 white Box II : 1 black 2 white and Box III : 2 black 2 white One of the box is selected at random and one ball is drawn. It turns out to be white. What is the probability of drawing white ball again, if the first one drawn is not replaced?

Answer 1

We're interested in the probability,

$$\mathbb{P}(2nd \;W | 1st \; W) = \frac{\mathbb{P}(1st \; W, 2nd \; W)}{\mathbb{P}(1st \; W)} $$

To find the denominator, condition on the which boxes where chosen,

\begin{array} \mathbb{P}(1st \; W) &=& \mathbb{P}(1st \; W | Box \; 1)\mathbb{P}(Box \; 1) + \mathbb{P}(1st \; W | Box \; 2)\mathbb{P}(Box \; 2)+\mathbb{P}(1st \; W | Box \; 3)\mathbb{P}(Box \; 3) \\ &=& \frac{1}{3}\frac{1}{3} + \frac{2}{3}\frac{1}{3} + \frac{2}{4}\frac{1}{3} \end{array}

To find the numerator, do same thing, except we're conditioning the event, $1st \; White$ AND $2nd \; White$, on the boxes chosen,

\begin{array} \mathbb{P}(1st \; W, 2nd \; W) &=& \mathbb{P}(1st \; W, 2nd \; W | Box \; 1)\mathbb{P}(Box \; 1) + \mathbb{P}(1st \; W, 2nd \; W | Box \; 2)\mathbb{P}(Box \; 2)+\mathbb{P}(1st \; W, 2nd \; W | Box \; 3)\mathbb{P}(Box \; 3) \\ &=& 0 + \frac{2}{3}\frac{1}{2}\frac{1}{3} + \frac{2}{4}\frac{1}{3}\frac{1}{3}\end{array}