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

Answer by K Sengupta)

We know that: Sum(i=0 to n) comb(n,i) = 2^i, and therefore:

the probability that precisely f flies, for 0<= f<= 5, are consumed by the spider = comb(5,f)/(2^f).

The spider will go hungry on a given day if at most 2 flies are consumed by it on that day, but will have eaten its fill it ift consumes at least 3 flies.

Accordingly, the probability that the spider will go hungry

= (comb(5,0) + comb(5,1) + comb(5,2))/(2^5)

= (1+5+10)/32

= 1/2

Similarly, the probability that the spider has eaten its fill

= (comb(5,3) + comb(5,4) + comb(5,5))/(2^5)

= (10+5+1)/32

= 1/2

If the spider has eaten its fill, the 6th fly will have a 100 percent probability of survival, but if the spider goes hungry, the probability that the fly will survive the attempt will be 1/2.

Consequently, the required probability that the 6th fly will survive the attempt to pass the web

= 1*(1/2) + (1/2)*(1/2)

= 3/4