# Probability of getting another white ball

There are 3 boxes. Each box has 2 balls. Box1: white, white. Box2: white, black. Box3: black, black.

You don’t know which box is box1, box2 or box3. All you did was getting a ball from a random box in front of you. And you got a white ball. What’s the probability of getting another white ball?

Please consider both situations: 1. You put the first white ball back to the box. 2. You discard the first white ball.

My thoughts: 1. First time you get a white ball 1/2 * second time you get a white ball 1/2 = 1/4 2. Only box 1 is possible for getting 2 white balls. So 1/3.

Are they correct? If not please explain.

• If you got a white ball initially, then you must have boxes 1 or 2, and is more likely to have been box 1. So the probability the next ball from the same box is white will exceed $\frac12$ (with or without replacement) – Henry Sep 27 '18 at 8:23

In the first case, the probability of picking Box 1 and drawing a white ball is $$\frac{1}{3}\cdot 1$$. The probability of picking Box 2 and drawing a white ball is $$\frac{1}{3}\cdot \frac{1}{2}$$. If you pick the black box, the probability to draw a white ball is 0. Hence, the probability to draw a white ball in setting 1. is $$\frac{1}{3}\cdot 1 +\frac{1}{3}\cdot \frac{1}{2} = \frac{1}{2}$$.
In setting 2 you have to take care of the box you picked in the first draw. If you drew a white ball from box 2, then your argument is correct. However, if in the first draw you drew a white ball from Box 1, then you could draw a second white ball from either box 1 or 2. You can draw twice from box 1 (probability $$\frac{1}{3}\cdot \frac{1}{3}$$) or from box 1 then from box 2 ($$\frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{2}$$) or from 2 then from box 1 ($$\frac{1}{3}\cdot \frac{1}{2}\cdot \frac{1}{3}$$). This adds up to $$\frac{1}{3}\cdot \frac{1}{3} + \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{2}+ \frac{1}{3}\cdot \frac{1}{2}\cdot \frac{1}{3} = \frac{2}{9}.$$