So it would be:

n = 5, p = 9/15 = 0.6 , q= 1-0.6 =0.4

P(x=3) = (5C3) * 0.6^3 *(0.4)^(5-3) = 0.3456

P(x=4) = (5C4) * 0.6^4 *(0.4)^(5-4) = 0.2592

P(x=5) = (5C5) * 0.6^5 *(0.4)^(5-5) = 0.07776

So adding the 3 prob = 0.68256 Ans.

**We will not consider n as 15. But n should be 5. Because we already have defined our selection i.e "of the first 5 databases". And n is the number of trials in a binomial distribution.**

**Am I right??**