Conditional Probability Problem

A black box contains 15 identical balls but in different color. They are 2 red and 13 black balls. (a) What is the probability that you grab a ball twice and they are both red if you never put the ball back into the box? (b) What is the probability that you get two red balls if you put the first ball back into the box and then grab the second one?


Solution:

(a) Ai = {the ith time grab a red ball}

   P(A1) = 2/15 
   (Because there are two red balls and totally 15 balls)
   P(A2|A1) = 1/14 
   (Because |A2∩A1| = 1 and there are 14 balls totally)
   P(A1∩A2) = (2/15) * (1/14) = 1/105

(b) Ai = {the ith time grab a red ball}

   P(A1) = P(A2) = 2/15
   P(A1∩A2) = (2/15) * (2/15) = 4/225

Comments/discussion

  • Where did you use conditional probability?
    • Answer here.
  • I think your problem involves independence. Do you know where? -pm


Back to first bonus point opportunity, ECE302 Spring 2013

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood