## Problem Involving Conditional Probability

Question:

If there are 2 white balls and 4 red balls in box 1, and 5 white balls and 3red balls in box 2. Now randomly take one ball from box 1 and put into box 2. Then take one ball from box 2.

- If under the condition that the ball being taken from box1 is red, what is the possibility that the ball that's taken from box 2 is also red?
- What's the possibility that the ball that's taken from box 2 is red?

Answer:

(1)Assume event A is the ball that's taken from box 2 is red; and event B is the ball that's taken from box 1 is red.

P(B) = 4/(2+4) = 2/3

P(A|B) =P(A_AND_B)/P(B)

= (2/3)*(4/9)/(2/3) =4/9

(2)

P(B) = 4/(2+4) = 2/3

P(NOT_B) = 1 - 2/3 = 1/3

P(A) = P(A_AND_B) + P(A_AND_NOTB)

= P(A|B)*P(B) + P(A|NOT_B)*P(NOT_B) =(4/9)*(2/3) + (1/3)*(1/3) =11/27

## Comments/discussion

- Write comment here
- answer here