Given three boxes X, Y, Z. Let W denote white balls and B denote black balls. The contents of the boxes are : X (2W, 3B) , Y(3W, 1B) , Z(1W, 4B).

Question

You need to select 1 box and from that draw 1 object at random. What is the probability that the object drawn is black ?

Approach 1 : Sample space = { (the box number, the ball) } // Sample Space = {(X,W) (X,W) (X,B) (X,B) (X,B) .... similar for Y and Z }

Thus answer is (3+1+4)/(2+3+3+1+1+4) = 8/14

Approach 2: summation ( probability of chosing ith box* prob of chosing a black) = (1/3 * 3/5) + (1/3 * 1/4) + (1/3 * 4/5) = 11/20

Which approach is correct and why ?

Answer 1

The second approach is correct. Just consider the extreme case you had the following setup:

Box X: (0W, 1B) <- no white balls, just 1 black ball

Box Y: (99W, 0B) <- loads of white balls, no black ball

Your first approach would give you a probability of 1% to get a black ball, but obviously, since you pick a box at random first and Box X does not contain white balls and Box Y does not contain black balls, the probability must be 50%. So it's

(1/2 * 1) + (1/2 * 0) = 1/2

Answer 2

Second Approach is correct.

There are basically 2 events , selecting the box and selecting the ball.

In first approach you are assuming that only 1 event is present(selecting the ball).