Probability and combination Problem?

Question

1. A bag contains 5 balls. Two balls are drawn without replacement and are found to be white. What is the probability that the bag contains 5 white balls? I know this is to be solved using Baye's therorem but i just can't define the events.

2. In how many ways can r flags of different colours be displayed on m poles (r may be <, > or =m)when there is no restriction on the number of flags that can be hoisted on one pole (i.e. it may range from 0 to r). What if each pole must have at least one flag (here r>= m).

Please show the steps of your solution.

Answer 1

To answer the first question, you need the prior probabilities. Consider, just for illustration, three extreme scenarios:
1: The 5 balls are known to have been drawn from a container with only 4 white balls so the probability of all five balls being white is zero. The two white balls don't change that.
2: The 5 balls are known to have been drawn from a container with only white balls so the probability of all five balls being white is one. The two white balls don't change that.
3: The a priori chance for any ball to be white is 1/2 and they are independent. So the probability of the remaining 3 balls to be white is (1/2)^3=1/8.

A more inteeresting scenario could be: A priori, the number of white balls could be 0, 1, 2, 3, 4, or 5, each with a probability of 1/6. Now the likelihoods of the two first balls to be both white are:
p|5=10/10
p|4=6/10
p|3=3/10
p|2=1/10

You see that the total probability mass is 20/10 so the answer to your question is (10/10)/(20/10)=1/2.

More generally, you could multiply the likelihoods with the prior probabilities q5, q4, q3 and q2:

p|5=10/10 * q5
p|4=6/10 * q4
p|3=3/10 * q3
p|2=1/10 * q2

So the general answer to your question is

q5
-----------------------------------------------------------
q5+6/10 q4+3/10 q3+1/10 q2

Answer 2

Problem 2. (Last updated again Tuesday sept. 5 03:10 EDT)
Problem 1:
A ball is either white or non-white (2 possible outcomes). But you have to know the statistical distribution of the 2 kind of balls. Let's suppose there is an equal number of each. If you have n balls in your bag then the probability of all them be white is 1 / (2^n). You take 2 white balls out so there is 3 balls remaining. The probability of all them being white is 1/(2^3) = 1/8 = 0.125
In the more general case the probability is the Expectancy (Average, mean) of a ball to be white raise to the third power (3 balls). The fact that you draw 2 white balls on a single event tells nothing about the probability of a ball to be white.

Problem 2.
On the first pole the number of possible ways is the sum of combinatorial from 1 to r in r plus one.
= 1 + Sum from i=1 to r (Combinatorial of i in r)
= 1 + Sum from i=1 to r ( r ! / ((r-i)! * i !))
The one is to take account of a possible zero number of flags on the pole. It would have to be taken out if you say there is minimum of one flag per pole.
On the second pole the number of possible ways is the sum of combinatorial from 1 r-a in r-a (where a is the number of flags on the first pole)
= Sum from i=1 to r-a (1 + Combinatorial of i in r-a)
= Sum from i=1 to r-a (1 + ( (r-a) ! / ((r-a-i)! * i !)))
So on for the third pole given that b flags were placed on the second pole
= Sum from i=1 to r-a-b (1 + Combinatorial of i in r-a-b)
= Sum from i=1 to r-a-b(1 + ( (r-a-b)! / ((r-a-b-i)! * i !)))
So the sum of all the possible combinations for all the poles will be the general case:

For m <= r
m + Sum from a = 1 to r-1 (
Sum from i=1 to r-a (1 + ( (r-a) ! / ((r-a-i)! * i !)))
)
The m factor added account for the poles where the last flag can go.

For m > r
(m! / ((m-r)! * r!)) + m + Sum from a = 1 to r-1 (
Sum from i=1 to r-a (1 + ( (r-a) ! / ((r-a-i)! * i !)))
)

The Combinatorial of m-r in m term added account for the number of ways the at least m-r poles without a flag can be arranged. The possibility of any other pole to be without a flag is accounted by the + 1 term in the most interior sum. I think this is the last update.