Solution to 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?

Answer 1

Two balls are certainly white. So the remaining balls may be one of the following
{WW, WN, NW, NN}
where W means a white ball and N means non-white

ball
Let the events be:
E- all four balls are white(WW)

F- 3 balls are white(WN,

NW)
G- only two balls are white(NN)

Now, P(E) =

P(G) =

1/4, and P(F) =

1/2
Let A be the event that the two balls drawn are white
P(A|E)

=

4C2/4C2 =

1
P(A|F)

=

3C2/4C2 =

1/2
P(A|G)

=

2C2/4C2 =

1/6
Now By Bayes' Theorem,
required probability =

P(E|A)

=

(1*1/4) / ( 1*1/4 + 1/2*1/2 + 1/6*1/4 )
=

(1/4) / ( 13/24 )
=

6/13
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2nd opinion

Let's say we have 3 boxes with 4 balls each.

Box A has 4 white balls.

Box B has 3 white balls.

Box C has 2 white balls.

The probability of drawing 2 W balls from;

Box A, P(2W│A) =

(4/4)∙(3/3) =

1

Box B, P(2W│B) =

(3/4)∙(2/3) =

1/2

Box C, P(2W│C) =

(2/4)∙(1/3) =

1/6

Say the probability of picking any of the 3 boxes is the same, we have;

P(A) =

1/3

P(B) =

1/3

P(C) =

1/3

Question is, given the event of drawing 2 W balls from a box taken blindly

from the 3 choices, what is the probability that the balls came from box A,

P(A│2W).

Recurring to Bayes Theorem:

P(A│2W) =

[P(A)P(2W│A)]/[P(A)P(2W│A)+P(B)P(2W│B)+P(C)P(2W│C)] =

[(1/3)(1)]/[(1/3)(1)+(1/3)(1/2)+(1/3)(1/6)] =

6/10 =

0.60 =

60%

P(A│2W) =

0.60 =

60%