Studying Probability Theory

Chapter 1.
How to prove that the r combination from n elements is n!/((n-r)!*r!) ?
Answer: we may first find out the permutation of r elements from n: we have n*(n-1)*(n-2)...*(n-r+1) = n!/(n-r)!. In these r elements permutation, if all elements are the same value of n, they belong to the same combination. And there are r! permutation of these r elements of the same values. So the combination is n!/(n-r)!/r!.

Chapter2.
Q1, when to use 全概率公式, when to use Bayes公式?
全概率公式 is used to calculate a single event probability. While Bayes公式 is used to calculate the conditional probability, that is we should first know one informational event and want to get another event probability. So we may define the known information event as B and getting event as A1.

14:
Since we know one information event that the target is destroyed, we may use Bayes theory. Let's define this event as B, and event A1, A2, A3 as the cannonball is shooted from 250m, 200m, 150m. We are seeking for P(A1|B).

P(A1|B)= P(A1)*P(B|A1)/(P(A1)*P(B|A1)+ P(A2)*P(B|A2)+ P(A3)*P(B|A3))
             = 0.1*0.05/(0.1*0.05+ 0.7*0.1+ 0.2*0.2)
             = 1/23

15:
Define the known event B to be receiving the ABCA and event A1, A2, A3 to be sending AAAA, BBBB, CCCC. So, we are seeking for P(A1|B).
Using Bayes theory, P(A1|B)= P(A1)*P(B|A1)/(P(A1)*P(B|A1)+ P(A2)*P(B|A2)+ P(A3)*P(B|A3)).
P(B|A1)= 0.6*0.2*0.2*0.6 = 0.0144
P(B|A2)= 0.2*0.6*0.2*0.2 = 0.0048
P(B|A3)= 0.2*0.2*0.6*0.2 = 0.0048

P(A1|B)= 0.5625

16:
P((AUB)C)=P((AC)U(BC))= P(AC) +P(BC) - P(ACBC)= P(A)P(C) + P(B)P(C) - P(A)P(B)P(C) =(P(A) + P(B) - P(A)P(B))P(C)
                  =P(AUB)P(C)
=> AUB is independent with C.

P(ABC)=P(A)P(B)P(C) =P(AB)P(C)   => AB is independent with C.

P((A-B)C)= P(A(~B)C)= P(AC(~B))= P(AC(omega - B))
                = P(AC) - P(ABC)
                = P(A)P(C) - P(A)P(B)P(C)
                = P(C)(P(A) - P(A)P(B))
                = P(C) P(A-B)
=> A-B is independent with C.

23:
The small probability event is A, so P(A)=p is very small.

In n-times Bernoulli trial, using binomial distribution, the b(0; n, p) = q^n. So the probability of A occurs is 1- b(0;n, p) = 1- q^n.

Since q<1, while n->infinite, q^n ->0 => 1- q^n -> 1. So follow the increase of n, the probability of A occurs will become 1, which means: small probability event will definitely occur for enough large n.

[Correction] This is not a Bernoulli trial.
[Correct Solution] The i-th times occurrence of the small probability event is A(i). If the small probability event did not occur in the first n times trial, the event is defined as B(n) = (~A(1))(~A(2))(~A(3))...(~A(n)). Since ~A(i) are independent, P(B(n))=P(~A(1))P(~A(2))P(~A(3))...P(~A(n))= (1-p)^n          

So ~B(n) means the small probability event will occur at least once in n times trial. P(~B(n))= 1 - (1-p)^n
When n->infinite, the P(~B(n)) will become 1. Thich means: small probability event will definitely occur for enough large n.

31:
Let's define event A: getting black ball from its package. p = P(A) = 1/N, q= 1-p =(N-1)/N
To switch back the black ball into the first package, event A must occur even times: 0, 2, 4, ..... We define this event as E. While the occurrence of odd times of A is defined as event O.
Using binomial distribuation:
P(E) = C(0, n)*p^0*q^n + C(2, n)*p^2*q^(n-2) + ...
P(O) = C(1, n)*p^1*q^(n-1) + C(3, n)*p^3*q^(n-3) + ...
=> P(E) + P(O) =1 and P(E) - P(O) = (q - p)^n = ((N-2)/N)^n
=> P(E) = 1/2+  ((N-2)/N)^n/2

1. N=1, while n=2k-1, P(E) = 0, n =2k P(E)=1
2. N=2  P(E)= 1/2
3. N>2  n -> infinite, P(E) ->1/2, note, P(E) >1/2