##### Contents

Combinatorics

Probability & Statistics

# Probability

### Binomial Distribution

```We have seen that the number of combinations of r elements
taken from a set of n elementsto is C(r,n) = n!/r!(n-r)!.

It is also the number os r-subsets built from a set of n elements.

It is also the number of ways of picking r unordered outcomes from n
possibilities

It is also the binomial coefficient. That is the purpose of this
chapter.

```

#### 1. Binomial coefficients

```Let's develop the binomial (a+b)2:
(a+b)2 = a2 + 2ab + b2
and (a+b)3
(a+b)3 = a3 + 3a2b + 3ab2 + b3

Let's develop (a+b)3 = (a+b)(a+b)(a+b) as follows:
(a+b1)(a+b2)(a+b3) = a3 + a2(b1+b2+b3) + a(b1b2 + b1b3 + b2b3) + b1b2b3
All the factors of a are combinations. For example in the
third term a(b1b2 + b1b3 + b2b3), the factor (b1b2 + b1b3 + b2b3)
gives all (three) the combinations to choose two "b"s from three,
their numer is: C(2,3) = 3. We can write:

(a+b)3 = C(0,3) a3 + C(1,3)a2b + C(2,3)ab2 + C(3,3) b3
or, more precisely,
(a+b)3 = C(0,3)a3b0 + C(1,3)a2b1 + C(2,3)a1b2 + C(3,3)a0b3

Now, we generalize to obtain the Newton binomial, that is
the expression of (a+b)n.

(a+b)n = C(0,n)anb0 + C(1,n)an-1b1 + C(2,n)an-2b2
+ ... + C(n,n)a0bn
or
(a+b)n = ∑ C(i,n)an-ibi				[i: 0 → n]

```

#### 2. Binomial distribution

```The probability P(x) that an event occurs x times among
n independent trials which the outcome of a single trial
is dichotomous (p ≡ success/ q = 1 -p ≡ failure)
takes the expression:
P(x) =  C(x,n)pxqn - x = (n!/x!(n - x)!) px(1 - p)n - x
This is the discrete Binomial distribution.

Binomial distribution:
P(x) =  C(x,n)pxqn - x = (n!/x!(n - x)!) px(1 - p)n - x

This distribution has:
1. The mean: μ = np
Proof:

x = ∑ P(x) x
where P(x) = (n!/x!(n - x)!) px qn - x
We have p ∂P(x)/∂p = P(x) x
Then:
x = ∑ P(x) x = ∑ p ∂P(x)/∂p
= p ∑ ∂P(x)/∂p = p ∂[∑ P(x)]/∂p
= p ∂[∑ (p + q)n]/∂p = pn (p + q)n - 1  = pn

2. The variance: σ2 = npq = np(1-p)
Proof:
Let m = n

σ2 =  (n - m)2
=  (n2 - 2nm + m2 )
= n2 - 2n m + m2 )
= n2 - m2 = n2 - (n )2

n2 = ∑P(x) x2 = (p∂/∂p)2 ∑P(x) = (p∂/∂p)(p∂/∂p) (p+q)n = (p∂/∂p) p (p+q)n - 1
= p [n(p + q)n-1 + pn (n-1)(p+q)n -2] = pn [1 + pn- p] = nn(q + pn)

σ2 = n2 - (n )2 =  pn (q + pn) - (pn) npq = npq

Binomial distribution: μ = np and σ2  =  npq

```

### 3. The Binomial distribution example

```
Let's recall that:
Binomial distribution:
P(x) =  C(x,n)pxqn - x = (n!/x!(n - x)!) px(1-p)n-x

Now, let's consider the following game:
We roll three six-sided dice (or one six-sided die three times); and we
consider the event E "obtain an ace". For one trial we have the probability
p =1/6 to get success and 1 - 1/6 = 5/6 to fail.

This binomial distribution has:
Mean value or expectation value μ = np
Variance or dispersion σ3 = np(1-p)

The probability to get x success among n trials is:
P(x) = (n!/x!(n - x)!) px(1 - p)n - x

We will consider the case of three trials. Hence (n = 3):
μ = 3 x 1/6 = 1/5
σ3 = np(1-p) = 1/2 x 5/6 = 5/12

P(x) = (3!/x!(3 - x)!) px(1 - p)3 - x

We have:
P(0) = (1 - 1/6)3 = (5/6)3 = 125/216
P(1) = 3 x (1/6) x (5/6)2 = 25/72 = 75/216
P(2) = 3 x (1/6)2 x (5/6) = 15/216
P(3) = 1 x (1/6)3 x 1 = 1/216
(We can verify that (p+q)3 = 13 = P(0) + P(1) + P(2) + P(3) = 1)

```

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