Vtyjkyuk

An urn contains 5 balls. 2 balls are drawn and are found to be white. What is Probability that all balls are white

$bfMy; Try::$ Let event $bfE_1=2$ drawn balls are white.

Let event $bfE_2=3$ drawn balls are white.

Let event $bfE_3=4$ drawn balls are white.

Let event $bfE_4=5$ drawn balls are white.

So Here $displaystyle bfbfP(E_1) = binom52;;;,P(E_2)=binom53;;;,P(E_3)=binom54;;;,P(E_4)=binom55;;;$

I have seems that Here we have used Bayes Theorem, But I did not understand How can we ued it.

Please help me,

Thanks

Since the two drawn are white, that means there were no more than 3 black balls in the original distribution. We have 2 choices for each ball color, so there are $2^5=32$ ways to assign the colors. But $binom55=1$ of those choices is all black and $binom54=5$ of those choices include exactly 4 black balls. Thus our sample space is reduced from 32 to
$$
32 - binom55 - binom54 = 32 - 1 - 5 = 26.
$$
Thus the probability is $dfrac126$ that all 5 are white.

The Universe of events in this problem is partitioned into
the mutually exclusive, and totally exhaustive, events $A_k$
denoting the fact that the urn contains $k$ white balls
$$
U = A_,,0 cup A_,,1 cup cdots cup A_,,5
$$

You don't specify the process by which the urns are filled-up
prior to the extraction, so we leave for the moment undefined.

Let $E_2$ denote the event "first two balls extracted are white".

Then given the urn $A_k$, with 5 balls of which $k$ white
the number of ways to extract the first two balls white, will
be equal to the number of ways to extract $k-2$ white balls from the remaining $3$:
$3 choose k-2$ over a total number of ways $5 choose k$, i.e.:
$$
Pleft( E_,,2 right) = left( matrix
3 cr
k - 2 cr right)/left( matrix
5 cr
k cr right) = left[ 2 le k right]3!k! over left( k - 2 right)!5! = k^,underline ,2, /5^,underline ,2, = 2! over 5^,underline ,2, left( matrix
k cr
2 cr right)
$$
where $x^,underline ,n, $ denotes the Falling Factorial.

Then
$$
eqalign
& Pleft( E_,,2 right) = sumlimits_k Pleft( E_,,2 right)Pleft( A_,,k right) = sumlimits_k Pleft( E_,,2 cap A_,,k right) = cr
& = 2! over 5^,underline ,2, sumlimits_0, le ,k, le ,5 left( matrix
k cr
2 cr right)Pleft( A_,,k right) = 1 over 5^,underline ,2, sumlimits_0, le ,k, le ,5 k^,underline ,2, Pleft( A_,,k right) cr
$$
and
$$
Pleft( E_,,2 cap A_,,k right) = k^,underline ,2, over 5^,underline ,2, Pleft( A_,,k right)
$$

Therefore we may conclude that
$$
Pleft( A_,,k right) = Pleft( A_,,k cap E_,,2 right) over Pleft( E_,,2 right) =
Pleft( E_,,2 cap A_,,k right) over Pleft( E_,,2 right) =
k^,underline ,2, Pleft( A_,,k right) over sumlimits_0, le ,k, le ,5 k^,underline ,2, Pleft( A_,,k right)
$$

Let now do some assumptions regarding the probability of urn composition.

1) all urns compositions equi-probable, with $P(A_k)=1/6$.

$$
Pleft( E_,,2 right) = 2! over 6;5^,underline ,2, sumlimits_0, le ,k, le ,5 left( matrix
k cr
2 cr right) = 2! over 6;5^,underline ,2, left( matrix
6 cr
3 cr right) = 2! over 3! = 1 over 3
$$
and
$$
eqalign
& Pleft( A_,,k right) = k^,underline ,2, Pleft( A_,,k right) over 5^,underline ,2, Pleft( E_,,2 right) = 1 over 2 cdot ,5^,underline ,2, k^,underline ,2, = 1 over 5^,underline ,2, left( matrix
k cr
2 cr right) = cr
& = left[ 0,;0,;1 over 20,;3 over 20,;6 over 20,;10 over 20 right] cr
$$

2) only urns with $2 le k$ equi-probable, with $P(A_k)=[2le k] 1/4$.

$$
Pleft( E_,,2 right) = 2 over 4;5^,underline ,2, sumlimits_2, le ,k, le ,5 left( matrix
k cr
2 cr right) = 1 over 2
$$
and we get the same result as above
$$
Pleft( A_,,k right) = k^,underline ,2, Pleft( A_,,k right) over 5^,underline ,2, Pleft( E_,,2 right) = left[ 2 le k right]k^,underline ,2, over 2;5^,underline ,2, = 1 over 5^,underline ,2, left( matrix
k cr
2 cr right)
$$

3) urns randomly filled with white/black balls, with $P(A_k)=5 choose k/2^5$.

$$
eqalign
& Pleft( E_,,2 right) = 1 over 2^,5 ;5^,underline ,2, sumlimits_0, le ,k, le ,5 k^,underline ,2, left( matrix
5 cr
k cr right) = 1 over 2^,4 ;5^,underline ,2, sumlimits_0, le ,k, le ,5 left( matrix
5 cr
k cr right)left( matrix
k cr
2 cr right) = cr
& = 1 over 2^,4 ;5^,underline ,2, left( matrix
5 cr
2 cr right)sumlimits_0, le ,k, le ,5 left( matrix
3 cr
k - 2 cr right) = 2^,3 over 2^,4 ;5^,underline ,2, left( matrix
5 cr
2 cr right) = 1 over 4 cr
$$
and
$$
eqalign
& Pleft( A_,,k right) = k^,underline ,2, over 2^,3 ;5^,underline ,2, left( matrix
5 cr
k cr right) = 1 over 2^,2 ;5^,underline ,2, left( matrix
k cr
2 cr right)left( matrix
5 cr
k cr right) = cr
& = 1 over 2^,2 ;5^,underline ,2, left( matrix
5 cr
2 cr right)left( matrix
3 cr
k - 2 cr right) = 1 over 2^,3 ;left( matrix
3 cr
k - 2 cr right) = cr
& = left[ 0,;0,;1 over 8,;3 over 8,;3 over 8,;1 over 8 right] cr
$$

Given that there are $n$ white balls in the urn we can find the probability of drawing two white balls.

$P(2 white balls chosen |n white balls in the urn) = frac nchoose 25choose 2$

However you have been asked to find $P(n white balls in the urn|2 white balls chosen )$

We can apply Bayes law.

$P(n|2) = P(2|n) frac P(n)P(2)$

But there is still not enough information to answer the problem.

We have to think about how the urn was filled and get into the head of the urn filler.

Scenario 1: When the urn was filled the urn filler rolled a die, and the die told him how many white balls to put into the urn. In this case before you started drawing $0$ white balls in the urn was as likely as $5$ white balls in the urn. After drawing 2, you know that there were not fewer than 2 white balls in the urn.

$P(5 white balls in the urn|2 chosen) = frac11+0.6+0.3+0.1=frac 12$

Scenario 2: The urn filler was blindfolded and filled the urn from a pool ofa balls that was 50/50 white and colored. 2 colored balls in the urn is more likely than 0 colored balls in the urn.

$P(5 white balls in the urn|2 chosen) = frac11+3+3+1=frac 18$

Scenario 3: The urn filler is a joker who always fills his urns with white balls.

$P(5 white balls in the urn|2 chosen) = 1$

And there are countless others.

Now you can take a weighted average based on the confidence you have in each of the assumptions, and arrive at your conclusion.

At this point there is no definitive right or wrong answer, as you are applying your own judgement.