I must be the worst person at math in the universe, oh well here goes.

My problem is that i don't understand the explanation (i don't know how to calculate it).

The probability question is as following:

You have 6 coins, what is the probability that you throw 4 times tails.

La place definition is the amount of beneficial outcomes divided by the amount of total possible outcomes.

so far so good.

The amount of total possible outcomes = 2*2*2*2*2*2 = 64

And here is where i stumble.

The description is (6)

(4) = 15 Beneficial outcomes

And i am completely lost

what do they mean by (6)

(4) = 15

And how to calculate it?

I am not sure that it is possible to explain even the rudiments of probability theory on a site like this. However, let's try to eliminate some darkness.

Some notation

FACTORIAL

4! = 4 * 3 * 2 * 1

3! = 3 * 2 * 1

2! = 2 * 1

1! = 1

A more formal definition is

\(\displaystyle 0! = 1\ and\)

\(\displaystyle n! = n * (n - 1)!\ for\ any\ n \in \mathbb N^+.\)

BINOMIAL COEFFICIENT

\(\displaystyle n,\ k \in \mathbb Z\ and\ 0 \le k \le n \implies \dbinom{n}{k} = \dfrac{n!}{k! * (n - k)!}.\)

Assume we flip a FAIR coin 6 times and write down the results as we get them.

As you saw there are 64 possibilities. And if the coin is fair, the probability of any one of them is

\(\displaystyle \dfrac{1}{64}.\)

I suggest that you write every possibility down in a systematic way, starting with HHHHHH and ending with TTTTTT. Count them up. You should have 64 and no duplicates.

How many cases will have 6 heads? Obviously 1. But

\(\displaystyle \dbinom{6}{6} = \dfrac{6!}{6! * (6 - 6)!} = \dfrac{1}{(6 - 6)!} = \dfrac{1}{0!} = \dfrac{1}{1} = 1.\)

How many cases will have 5 heads. It is not so obvious, but it is 6 cases because the T could come first, second, third, fourth, fifth, or sixth. And

\(\displaystyle \dbinom{6}{5} = \dfrac{6!}{5! * (6 - 5)!} = \dfrac{6 * 5 * 4 * 3 * 2 * 1}{5 * 4 * 3 * 2 * 1 * 1!} = \dfrac{6}{1} = 6.\)

How many cases will have 4 heads. I suggest you count them up. You will find if you have not made a mistake that you have 15. They will look like this:

HHHHTT 1

HHHTHT 2

HHTHHT 3

HTHHHT 4

THHHHT 5

HHHTTH 6

HHTHTH 7

HTHHTH 8

THHHTH 9

HHTTHH 10

HTHTHH 11

THHTHH 12

HTTHHH 13

THTHHH 14

TTHHHH 15

And

\(\displaystyle \dbinom{6}{4} = \dfrac{6!}{4! * (6 - 4)!} = \dfrac{6 * 5 * 4 * 3 * 2 * 1}{ 4 * 3 * 2 * 1 * 2!} = \dfrac{6 * 5}{2 * 1} = 15.\)