Question

A box contains 80 balls numbered from 1 to 80. If 15 balls are drawn with replacement, what is the probability that at least two of them have the same number?

Answer #1

Answer)

As there are fixed number of trials and probability of each and every trial is same and independent of each other

Here we need to use the binomial formula

P(r) = ncr*(p^r)*(1-p)^n-r

Ncr = n!/(r!*(n-r)!)

N! = N*n-1*n-2*n-3*n-4*n-5........till 1

For example 5! = 5*4*3*2*1

Special case is 0! = 1

P = probability of single trial = 1/80 = 0.0125

N = number of trials = 15

R = desired success = at least 2

We know that sum of all the probabilities is = 1

So, P(at least 2) = 1 - (P(0) + p(1))

= 0.01472530971

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