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?

Let A denote the event that the sixth fly survives and B denote the event that the spider has already had his quota of 3 flies filled from the first five flies. Let C denote the complementary event for B, i.e., that the spider still needs one or more flies for his quota.
We need to find P(A).
P(A)=P(A and B) + P(A and C)
=P(A|B).P(B)+P(A|C).P(C)
We are told that P(A|C) = (1/2). We assume that
P(A|B) is 1 since the spider is already satiated; ignoring the possibility that the spider could very well kill 'for pleasure' even if it is already full.
P(C)=P(spider caught 0,1,or 2 flies out of first 5)= (1+5+10)/32=1/2.
P(B)= 1-P(C)=1/2
So P(A)=1.(1/2)+(1/2).(1/2)=(3/4)

Posted by Prab on 2003-11-28 00:30:00