# Is my approach correct for this binomial distribution question?

#### sallu110

##### New Member
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??

#### staassis

##### Active Member
The original question does not display.

#### sallu110

##### New Member
The original question does not display.
A search engine randomly checks into 15 different databases. Only the 9 databases have the required
results. What is the probability that the required result will be found in at least 3 of the first 5 databases?

#### staassis

##### Active Member

[(9 choose 3) * (6 choose 2) + (9 choose 4) * (6 choose 1) + (9 choose 5) * 1] / (15 choose 5).

This is not a case of binomial distribution because the databases are sampled without replacement.

#### James Howard

##### New Member
is there an analytical formula for the PARTIAL sum for the binomial coefficient (n choose k) with respect to n?
e.g what is the sum(n choose k)) from n=0 to some integer N?

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