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# A spider eats 3 flies a day. Till the spider eats the three flies, any fly has a 50% chance of survival if it attempts to pass the web. What is the probability that the 6th fly attempting to pass will survive the attempt?

Yash Chourasiya
10 months ago

Hello Student.

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)!).

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

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

c)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 .