Here is the question:https://ibb.co/TLPVTmz
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??
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??