Guys,

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,

Then guys....................

what is the probability that the 6th fly will survive the attempt???

Its an easy one..... Try,

Thanks,
Riju.

Well The answer is 0.75. One needs to use Binomial Distribution to find out this.

Hi Friends,

Yes, The Answer is 0.75.

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 .

Congrats vcite..

Regards,
Riju

Thankyou Rijus.Good work

I guess, probability of the 6th fly of survival should be again 50%.