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.

  1. 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?
  2. 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

Back to first bonus point opportunity, ECE302 Spring 2013

Alumni Liaison

Prof. Math. Ohio State and Associate Dean
Outstanding Alumnus Purdue Math 2008

Jeff McNeal