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?
This didn't seem like a hard question, then I saw the disparity of the replies and gave it a second look. Since each individual fly has a 50% chance of survival, then the possibilities are that all survived (1/25) that one was eaten (5/25 that is A or B or C or D or E) that two were eaten (10/25 that is A and one of the other 4, or B and one of the other 3, or C and one of the other 2, or D & E) or that three were eaten (9/25 that is A B and one of the other three, A C and one of the other two, A D E, B C and one of the remaining two, or C D E). So in 16 of the 25 cases, he has a 50% chance of survival, or 16/25*.5 which equals 32%
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