Problem 169 Solution

Using the binomial distribution, the probability that n flies out of 5 have been eaten is combin(5,n)*(1/2)⁵, where combin(5,2)=5!/(n!*(5-n)!).

The probability that 0 flies were eaten is combin(5,0)*(1/2)⁵=1/32.

The probability that 1 fly was eaten is combin(5,1)*(1/2)⁵=5/32.

The probability that 2 flies were eaten is combin(5,2)*(1/2)⁵=10/32.

So the probability that the spider is still hungry is 1/32 + 5/32 + 10/32 = 16/32 = 1/2. The probability the spider is full is 1-1/2=1/2. Thus the probability of a successful attempt to pass is (1/2)*1 + (1/2)*0.5 = 0.75 .

Michael Shackleford, ASA - August 10, 2001