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?
(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
re: simpler computation
