11. The Theorem of total probability.

If B A then P(B) ≤ P(A).

Proof. A = B (A \ B), so

P(A) = P(B) + P(A \ B) P(B):

Def. The events A₁….A_n form a partition of the

sample space Ω if

1. A_i are mutually exclusive: A_i A_j = for i ≠ j.

2. A₁ …. An = Ω.

Total Probability Theorem. Let A_1…. A_n be

a partition of Ω. For any event B,

Proof. B= (B Aj) (disjoint union), so

The theorem follows from P(B Aj) = P(Aj) P(Bj|Aj).

The latter holds for Aj with Pr(Aj) = 0 if we define

P(Aj) P(Bj|Aj) := 0 since then P(B Aj) = 0

Example. In a certain county

60% of registered voters are Republicans

30% are Democrats

10% are Independents.

When those voters were asked about increasing military spending

40% of Republicans opposed it

65% of the Democrats opposed it

55% of the Independents opposed it.

What is the probability that a randomly selected voter

in this county opposes increased military spending?

P(Bj|R) = 0.4, P(Bj|D) = 0.65, P(Bj|I) = 0.55.

By the total probability theorem:

P(B) = P(Bj|R) P(R) + P(Bj|D) P(D) + P(Bj|I) P(I)

= (0.4 * 0.6) + (0.65 * 0.3) + (0.55 * 0.1) = 0.49.

12. Bayes’ Rule.

Bayes Theorem. Let A₁ ,….,A_n be a partition of

Ω. For any event B

Proof.

Example.

A registered voter from our county writes a letter

to the local paper, arguing against increased military spending. What is the probability that this

voter is a Democrat?

Presumably that is P(Dj|B), so by Bayes’ theorem:

P(Dj|B) =0.65*0.3\((0.4 * 0.6) + (0.65 * 0.3) + (0.55 * 0.1))

=0.195\0.49= 0.398

13. Bernoulli scheme. Bernoulli distribution.

Def1The number m of success in n Bernuli trails is called a binomial distribution. And its value will be denote by P_n(m) since the depend of a number of trails and the probability of success a given trail.

We wish to find a formula that gives the P_n(m) in n trails for binomial experiment p-number success probability, q=1-p- number failure probability. P^m*q^n-mC_n^m=n!\m!(n-m)!

Def2. A Bernuli trails can result in a success with probability p and a failure with probability q=1-p. then the probability distribution of binomial random variable m. then the number of success in n independent trails P_n(m)= C_n^m *P^m*q^n-m

Remark. The binomial sum

Def3. The mean and variance of binomial distribution and .

14. Poisson approximation formula.

Def1.Poisson distribution is called the distribution have experiment in interval for example minute, day, year, month, week and etc.

Def2. The probability distribution of Poisson random variable in representity the number of outcomes occurring in a given time interval or specify region denoted by t.

Where ℷ- is the average number of outcomes, n=0,1,2,3….∞ ℷ=np

Proof. P_n(m)= C_n^m *P^m*q^n-m- binomial distribution  P_n(m)= *p^m*(1-p)ⁿ*(1-p)^-m=(1-(1-1\n)* )*( )^m(1- )^m(1- )^-m= =1; = -1

ⁿ= ^1\ℷt)^-ℷt=e-^ℷt

^-m=(1) ^–m=1 

Remark. Poisson distribution sums is equal to