Question

For tossing a coin 20 times and and getting 4 head as an outcome... how do we get 4845?

Answer 1

4845 is `((20),(4))` and is the number of ways 4 Heads and 16 Tails can be achieved

Explanation:

In the binomial probability, we have the relationship:

`sum_(k=0)^(n)C_(n,k)(p)^k(1-p)^(n-k)=1`

What this says is that if we take all the possible situations in a given binomial probability situation, such as with flipping a coin a certain number of times, we'll end up with a probability of 1 - or in other words, we've accounted for each and every potential situation.

So let's take the situation where we flip a coin 20 times and sum up all those possibilities. And to start with, let's look at the situation where all the flips are Heads (and let's have Heads be "successful" so we set `P("Heads")=p=.5`). So we have HHHHH...HHH. 20 of them. There's only 1 way to achieve 20 heads, and so when looking at the term for all Heads, we have:

`((20),(20))(.5)^20(.5)^0=1xx.5^20xx1=` something very small (roughly `1xx10^-6`).

Now let's look at what happens with 19 Heads and 1 Tail. How many different ways can we achieve 1 Tail and 19 Heads?

THHH...HHH
HTHH...HHH
HHTH...HHH

and so on. There are 20 different ways to do it. So we have:

`((20),(19))(.5)^19(.5)^1=20xx.5^19xx.5=` something not quite as small but still small (roughly `0.00002`).

Let's move now to your question - we have 4 Heads. How many ways can we get 4 Heads and 16 Tails? That's:

`((20),(4))=(20!)/((4!)(16!))=(20xx19xx18xx17)/24=4845` different ways to have 4 Heads and 16 Tails - and so many more ways to have happen than all Heads. Like 4845 times more ways.

The full binomial is:

`((20),(4))(.5)^4(.5)^16=4845xx.5^4xx.5^16~~0.0046`