What is the relationship between binomial theorem and binomial probability distribution?

Question

I should be able to figure this out, but I can't. I'm having a difficult time figuring out the binomial theorem, itself.

Answer 1

The binomial theorem says that when you have:

(x+y)^n, then when you expand this out (multiply it out) that terms will be of the form:

(x^(n-k))*(y^k), and the coefficient (number in front) of this term will be equal to C(n,k), that is the amount of combinations when you pick out "k"  number of items from a basket on "n" number of items.  k represents just the kth term in the series.  n will be given as a number.  

Here's an example of how it relates to probability distribution:

Find the probabilty that when i roll 6 dice, that a 6 never occurs out of these six rolls.  

This can be represented by:

P(never a 6) = P(1,2,3,4,5)*(P(1,2,3,4,5)*P(1,2,3,4,5)....

where P(1,2,3,4,5) means the probability of getting a 1,2,3,4, or 5 on a single roll.  And this is equal to 5/6.  

so P(never a 6) = (5/6)*(5/6)*(5/6)....(5/6)     =  (5/6)^6

But what if we asked a different question, what is the probability that a 6 does occur somewhere in the 6 rolls??

it would be simply the complement of the probabilty that a 6 never occurs.  So the probability would be 1-(5/6)^6.

However, we could do this the hard way too.  In other words, the probabiltity that a six does occur somewhere in six rolls is equal to:

P(6 occurs just once) + P(6 occurs just twice) + ..... + P(6 occurs all 6 times).  

Out of 6 rolls, there are six possiblites of rolling a 6 just once.  It can be rolled on the first roll, on the second roll, third roll, fourth roll, fifth roll, or sixth roll.  So there are six combinations, and we multiply that by 1/6 (the probability of getting a six).  However, if one roll is six then all the others must not be six (because its the probability that it just occurs once).  So we must multiple that by 5/6 to the 5th power.

= 6*(1/6)*(5/6)^5.

What about getting just 2 sixes out of the 6 rolls?

Well, there are 15 combinations at which his can occur.  The amount of combinations of picking two die out of a pile of six.  And we multiply this by (1/6)^2 (probability of getting two sixes on two rolls).  If 2 dice are 6's then the rest must not be (probability of JUST getting two sixes).  Therefore, we multiply this by (5/6)^4 and the result is:

15*(1/6)^2*(5/6)^4

we add this result to the previous result.  

We keep doing this until we find the probability of getting six 6's in 6 rolls.  

The terms are thus:

6*(1/6)*(5/6)^5

15*(1/6)^2*(5/6)^4

20*(1/6)^3*(5/6)^3

15*(1/6)^4*(5/6)^2

6*(1/6)^5*(5/6)

(1/6)^6

We add all 6 terms together, and we get the result of the probability of rolling a six.  

Now what if we said we wanted the probability of rolling a six or never rolling a six out of the six rolls.  It would simply be one.  And the probability of never rolling a six was found before to be (5/6)^6.  If we add this term to the terms that we found that represent rolling atleast a six, then we should get one.  So we get:

(5/6)^n + 6*(1/6)*(5/6)^5 + 15*(1/6)^2*(5/6)^4 + 20*(1/6)^3*(5/6)^3 + 15*(1/6)^4*(5/6)^2 + 6*(1/6)^5*(5/6) + (1/6)^6 = 1

and 1 can be represented as (5/6 + 1/6)^6

so (5/6 + 1/6)^ 6 can be expanded to

(5/6)^6 + 6*(1/6)*(5/6)^5 + 15*(1/6)^2*(5/6)^4 + 20*(1/6)^3*(5/6)^3 + 15*(1/6)^4*(5/6)^2 + 6*(1/6)^5*(5/6) + (1/6)^6

which is precisely the binomial theorem.