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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?

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?

Grade:12th pass

1 Answers

Yash Chourasiya
askIITians Faculty 256 Points
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 .


I hope this answer will help you.

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