All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Probability
Surprising answer (Posted on 2015-09-02) Difficulty: 3 of 5
Suppose we put eight white and two black balls into a bag and then draw forth the balls one at a time.
If we repeat this experiment many times, which draw is most likely to produce the first black ball?

Try to guess (=estimate guided by intuition), then evaluate.
You'd be surprised.

Source: A.E. Lawrence, “Playing With Probability”
Mathematical Gazette, (1969)

No Solution Yet Submitted by Ady TZIDON    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solution | Comment 1 of 6
Probability that the first black ball occurs at

Draw 1: 2/10 = 1/5 = 0.2 (the most likely)
Draw 2: 4/5 * 2/9 = 8/45 = 0.17777...
Draw 3: 4/5 * 7/9 * 2/8 = 7/45 = 0.155555...
Draw 4: 4/5 * 7/9 * 6/8 * 2/7 = 2/15 = 0.1333333...

continuing:

Draw 5: 1/9 = 0.11111...
Draw 6: 4/45 = 0.08888...
Draw 7: 1/15 = 0.06666...
Draw 8: 2/45 = 0.04444...
Draw 9: 1/45 = 0.02222...


So the modal case is not the mean case. I guess that's the surprise. It's less surprising if there were replacement, as then it would be obvious that each successive delay just requires more  successive white ball draws. Here, you might think the successively larger proportion of black balls might offset the requirement of larger numbers of white ball choices, but that's not the case.

  Posted by Charlie on 2015-09-02 11:45:16
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (9)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information