Combinatorics - Probability

Probability with two outcomes
Binomial theorem

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Bernoulli experiment (binomial experiment)

Let's consider the following lottery books:
A is the event associated to "win a book", and B is the event associated to "not win a book"

P(A) = p
P(B) = 1 - p = q

The events A and B are mutually exlusive.(they cannot both exist at the same time). The lottery consists of 4 trials

1. There is ONE WAY to succeed (win) all of the four trials (four trials om the four trials):
P(to Succeed (win) all of the four trials) =
PS(4/4) = p x p x p x p = p⁴
ONE WAY = C(4,4) = 4!/4!0! = 1
PS(4/4) = p x p x p x p = p⁴
Then:
PS(4/4) = C(4,4) p⁴q⁰

2. There are C(3,4) = 4!/3!1! = 4 WAYS to succeed (win) three of the four trials:
1. [S,S,S,F] : p x p x p x q = p³q¹
2. [S,S,F,S] : p x p x q x p = p³q¹
3. [S,F,S,S] : p x q x p x p = p³q¹
4. [F,S,S,S] : q x p x p x p = p³q¹

Then:
P(3/4) = C(3,4)p³q¹

3. There are C(2,4) = 4!/2!2! = 24/4 = 6 WAYS to succeed (win) two of the four trials:
1. [S,S,F,F] : p x p x q x q = p²q²
2. [S,F,S,F] : p x q x p x q = p²q²
3. [F,S,S,F] : q x p x p x q = p²q²
4. [S,F,F,S] : p x q x q x p = p²q²
5. [F,F,S,S] : q x q x p x p = p²q²
6. [F,S,F,S] : p x q x q x p = p²q²

Then:
P(2/4) = C(2,4)p²q²

4. There are C(1,4) = 4!/1!3! = 4 WAYS to succeed (win) one of the four trials:
1. [S,F,F,F] : p x p x q x q = p¹q³
2. [F,S,F,F] : p x q x p x q = p¹q³
3. [F,F,S,F] : q x p x p x q = p¹q³
4. [F,F,F,S] : p x q x q x p = p¹q³

Then:
P(1/4) = C(1,4)p¹q³

5. There are C(0,4) = 4!/0!4! = 1 WAY to succeed (win) zero of the four trials:
1. [F,F,F,F] : q x q x q x q = q⁴

Then:
P(0/4) = C(0,4)p⁰q⁴
(= P(to Fail (lose) all of the four trials) = PF(4/4) = (1 - p)⁴)

The probability to win the lottery books is:
PS = PS(4/4) + P(3/4) + P(2/4) + P(1/4) + P(0/4) =
C(4,4) p⁴q⁰ + C(3,4) p³q¹ + C(2,4) p²q² + C(1,4) p¹q³ + C(0,4) p⁰q⁴
= Σ C(i,4) pⁱq^{(n - i)} = (p + q)ⁿ = (p + 1 - p)ⁿ = 1.

Example

What is the probability to win two books, the first at the second trial and the second at the last trial?
This is the sixth way of the third case :
"3.There are C(2,4) = 4!/2!2! = 24/4 = 6 WAYS to succeed (win) two of the four trials:
6. [F,S,F,S] : p x q x q x p = p²q²
"
P(2/4) = C(2,4)p²q² = 6 x p²q²
If p = q = 1/2, then:
P(2/4) = 6 x (1/2)²(1/2)² = 6 /16 = 3/8 = 37.5%.

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C(m,n) = n!/m!(n - m)!

The i^th term of the Newton's binomial is C(i,n) pⁱ q^{(n - i)}

p + q = 1: p for success, then 1 - p for failure

C(m.n) p^m q^{(n- m)} ways to succeed m times among n trials

Bernoulli experiment = binomial experiment: the events are mutually exlusive and independent

Newton's binomial:
(a + b)ⁿ = Σ C(m,n) p^m q^{(n - m)}
m: 0 → n

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