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 Spider and Fly (Posted on 2003-11-25)
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)

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 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

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