This is very easy.
I am stat PhD but this is very basic question.
Let me explain.
For the combination part, 9 choose 6, think of it in the following way.
We are dealing with a BINOMIAL DISTRIBUTION. What is that? This is the form when we have a trial such as flipping a coin with ONLY two possible outcomes. We have a head and a tail. So in statistical terms, this is a Bernoulli trial.
Now we are not just flipping the coin once. Are we? We are flipping the coin 9 times.
Now out of these 9 times, there are 6 outcomes that we want as tails. Right? However out of these nine slots, we could have these six tails appear in the first six, the last six, or interspersed throughout. Furthermore, we are doing this without replacement, hence we need to have 9 nCr 6 (9 choose 6) in front.
Now, let us define a "success" according to what we are interested in. and a "failure" according to what we are not interested in. Here the probability of a "success" (a tail in this case) is 1/2 (assuming the coin is unbiased). The probability of the failure is 1 - success = 1 - 1/2 = 1/2 which is the prob of the head. Now we raise the first half (p, the success) to the 6th power because this is the number of success. And we raise the probability of the failure (heads) the other half raised to the 3rd (9-6 failures)
So we have (9 choose 6) * (1/2)^6 * (1-1/2)^3
This is binomial distribution in the form.
f(x) = (n choose x) * p^x * (1-p)^ (n-x)
where n is the total number of trials. x is the number of success.
p is the probability of success. and 1-p is the probability of failure
f(x) is the probaiblity density function. That is, a form in which a certain distribution takes.
So understand the concept and just memorize the binomial formula for a problem like this, when there are only two outcomes, a success and a failure repeated multiple times.
