# Probability question - Bayes' Theorem - 4 Balls, 2 white, P(all white) = ?

dharavsolanki

## Homework Statement

A Bag contains 4 balls.two balls are drawn at random,and are found to be white.what is the probability that all balls are white?

## Homework Equations

P(B/A) = P(A intersection B) / P (A)

## The Attempt at a Solution

A = Two balls are white.
B = All four balls are white.

I wrote P(B/A) and P(A/B) according to Bayes' theorem and eliminated the intersection term, to arrive with P(B/A) = P(B)/P(A). Ofcourse, P(A|B) = 1.

How do i solve these two now?

I found an alternate solution on net which goes this way...
P(E1)=p(E2)=p(E3)=1/3.
Calculate p(A1),p(A2),p(A3), you get 1, 1/2, 1/4

Apply Baye's theorem and you get the magical 3/5 as the answer

and another which says...
p(white)=1/2
p(nonwhite)=1/2
assuming equally likely outcomes
given balls 1&2 are white their respective proababilities of being white is 1.and is 1/2 for the other two.
hence,
P(all are white)=p(first ball being white)*p(2nd being white)*p(3rd being white)*p(4th being white)
=1*1*1/2*1/2
=1/4

Can anyone, apart from helping me with my question, explain the first aliter and the mistake in assuming the probability = 1/2 in the second aliter?

## Answers and Replies

Staff Emeritus
Homework Helper
I don't think you can solve this without more information. For example, if all you have is white balls to fill the bag, then the probability is obviously 1. On the other hand, if white is just one of an equally likely million colors, the probability will be close to 0.

Homework Helper
Gold Member
Dearly Missed
Hi, dharavsolanki!

You have three possible states that two whites out of four can arise from:

A) 2 whites and 2 non-whites.
B) 3 whites and 1 non-white
C) 4 whites.

You are to determine the probability of C, given that you pick two whites.

The easiest way to solve this is as follows:

If you have case A), there will be a total of 4*3=12 possible outcomes (permutations included), with 2 of those outcomes being to get two whites.

If you have case B), there will also be 12 possible outcomes, with 3*2=6 of those being to get two whites.

If you have case C) there will also be 12 possible outcomes, ALL of which yield two whites.

Given that you get two whites, there is in total 2+6+12=20 outcomes you have picked 1 from!

Thus, the chance that this outcome lies in C) is simply 12/20=3/5.

willem2
If you have case A), there will be a total of 4*3=12 possible outcomes (permutations included), with 2 of those outcomes being to get two whites.

If you have case B), there will also be 12 possible outcomes, with 3*2=6 of those being to get two whites.

If you have case C) there will also be 12 possible outcomes, ALL of which yield two whites.

Given that you get two whites, there is in total 2+6+12=20 outcomes you have picked 1 from!

Thus, the chance that this outcome lies in C) is simply 12/20=3/5.

This is only right if you assume that the the a priori probabilities of cases A, B and C are all 1/3. That fact should have been given in the question.
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