# Thread: Discrete probobility distributions

1. ## Discrete probobility distributions

Okay, so my book is horrible and it doesnt explain things very well and i cant seem to find my answer anywhere online.

I wont write out the example in the book since i think its irrelevant and way to long. I am struggling with the notation and how they get the answer.

$\displaystyle f(0) = P(X=0) = \frac{\left[\begin{array}{cc}3 \\0 \\\end{array}\right] \left[\begin{array}{cc}17 \\2 \\\end{array}\right]}{\left[\begin{array}{cc}20 \\2 \\\end{array}\right]} = \frac{68}{95}$

I m very confused on how they got the answer.
I think the (3 0 ) = 0?

Can someone please explain and
work this out because i am completely lost and i am sure its easy.
I am wiling to look at a webpage if i knew what to look up.

Thank you

2. ## Re: Discrete probobility distributions

Originally Posted by icelated
Okay, so my book is horrible and it doesnt explain things very well and i cant seem to find my answer anywhere online.
I wont write out the example in the book since i think its irrelevant and way to long. I am struggling with the notation and how they get the answer.

$\displaystyle f(0) = P(X=0) = \frac{\left[\begin{array}{cc}3 \\0 \\\end{array}\right] \left[\begin{array}{cc}17 \\2 \\\end{array}\right]}{\left[\begin{array}{cc}20 \\2 \\\end{array}\right]} = \frac{68}{95}$
I m very confused on how they got the answer.
I think the (3 0 ) = 0?
Your textbook may well be 'horrible' but so is the way you have posted this question.
The fact that you chose not to post the actual question is horrible.

First of all, I assume that those 'square brackets' are really meant to be binomial coefficients.
But note that $\displaystyle \binom{K}{0}=1,~K\in\mathbb{N}~.$.

So what is the actual question?