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?
Given that the spider has not yet reached his quota, the fly will have a 50% chance of getting through... but we don't know whether or not the spider has reached his quota from the first 5 flies...
So, there are several possibilities that the spider could have reached his quota within the first 5 flies:
(let C = caught, and N = not caught)
Here they are, along with the likelihood of them occuring:
CCC = 1/8 = 4/32
CCNC = 1/16 = 2/32
CCNNC = 1/32
CNCC = 1/16 = 2/32
CNCNC = 1/32
CNNCC = 1/32
NCCC = 1/16 = 2/32
NCNCC = 1/32
NCCNC = 1/32
NNCCC = 1/32
These are ALL the possibilities (I think) of the spider having caught his fill for the day.
Now, if we total this up, we show there is a 1/2 chance that the spider has already caught his fill.
If he has, then he won't attempt capturing the fly.... and if he hasn't, there is a 50% chance he'll capture the fly...
The question asked... what's the probability the fly will make it... so
1/2 * 50% + 1/2 * 100% =
3/4
solution
