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

Home > Probability
Spider and Fly (Posted on 2003-11-25) Difficulty: 3 of 5
A spider eats 3 flies a day. Until the spider fills his quota a fly has a 50% chance of survival if he attempts to pass the web.

Assuming 5 flies have already made the attempt to pass, what is the probability that the 6th fly will survive the attempt?

See The Solution Submitted by Ravi Raja    
Rating: 3.6250 (8 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: simpler computation | Comment 7 of 21 |
(In reply to simpler computation by Charlie)

Charlie... I thought of that when I first looked at the problem.... but it is misleading.

Though your calculation 'works out', the spider NEVER eats 5 or 4 flies in the first 5 flies. Once he's eaten the first 3, he stops capturing them.

If you want to use combinatorial mathematics (and I purposefully, avoided the combination function for ease of understanding), you should total up the likelihood of getting 3 eaten if the THIRD fly is eaten, the likelihood of getting 3 eaten if the FOURTH fly is eaten, and the likelihood of getting eaten if the FIFTH fly is eaten....

This would correspond to:

1/8 chance:
CCC

1/16 chance (each):
CCNC
CNCC
NCCC

1/32 chance (each):
CCNNC
CNCNC
CNNCC
NCCNC
NCNCC
NNCCC

total: 1/2
_____________

All the 16 remaining possibilities don't have 3 flies eaten (and all have 1/32 chance of occuring)
NNNNN
NNNNC
NNNCN
NNCNN
NCNNN
CNNNN
CNNNC
CNNCN
CNCNN
CCNNN
NCNNC
NCNCN
NCCNN
NNCNC
NNCCN
NNNCC
total: 1/2

  Posted by SilverKnight on 2003-11-25 10:10:59

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 (12)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information