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A bag contain 4 balls 2 balls are drawn at random and are found to be white What is the probability that all balls are white?
it has three cases 1st E1- two balls are white
E2- three balls are white
E4-four balls are white
E4/w= (p(E3)*4C2/4C2)/(P(E1)*2C2/4C2+P(E2)3C2/4C2+P(E3)*4C2/4C2)
WHERE-P(E1)=P(E2)=P(E3)=1/3
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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%
What else can I help you with?
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