Question 430559: There are two sets of marbles. Both have 1,000 in them. Bag 1 has 700 blue and 300 red. Bag 2 has 300 blue and 700 red. Picking a bag at random you draw out 10 marbles with replacement. If 7 are red and 3 are blue, what is the exact probability of you picking Bag 1?
Found 2 solutions by Edwin McCravy, robertb:
Answer by Edwin McCravy(20056)   (Show Source):
You can put this solution on YOUR website! There are two sets of marbles. Both have 1,000 in them. Bag 1 has 700 blue and 300 red. Bag 2 has 300 blue and 700 red. Picking a bag at random you draw out 10 marbles with replacement. If 7 are red and 3 are blue, what is the exact probability of you picking Bag 1?
P(bag1 & 7R3B)
P(bag1 given 7R3B) = ----------------
P(7R3B)
P(7R3B|bag1) = C(10,3)[(.7)^7(.3)^3]
P(7R3B|bag2) = C(10,3)[(.3)^7(.7)^3]
Numerator = P(bag1 & 7R3B) = P(bag1)*P(7R3B|bag1) = (.5)C(10,3)(.7)^7(.3)^3
Denominator = P(7R3B) = P[(bag1 & 7R3B) OR (bag2 & 7R3B)] =
(.5)C(10,3)[(.7)^7(.3)^3] + (.5)C(10,3)[(.3)^7(.7)^3]
Numerator = P(Bag2 & 7R3B) = P(Bag2)*P(7R3B|Bag2) = (.5)C(10,3)(.3)^7(.7)^3
(.5)C(10,3)(.7)^7(.3)^3
-----------------------------------------------------
(.5)C(10,3)[(.7)^7(.3)^3] + (.5)C(10,3)[(.3)^7(.7)^3]
Divide top and bottom by (.5)C(10,3)
(.7)^7(.3)^3
-----------------------------
(.7)^7(.3)^3 + (.3)^7(.7)^3
Divide the top and bottom by (.7)^3(.3)^3
(.7)^4
---------------------------
(.7)^4 + (.3)^4
Multiply top and bottom by 10^4
7^4
-------------
7^4 + 3^4
2401
-------------
2401 + 81
2401
------
2482
Edwin
Answer by robertb(5830)  (Show Source):
You can put this solution on YOUR website! The assumptions that
P(7R3B|bag1) = C(10,3)[(.7)^7(.3)^3]
P(7R3B|bag2) = C(10,3)[(.3)^7(.7)^3]
are wrong.
The experiment of drawing balls from each bag doesn't follow a binomial distribution, but a hypergeometric distribution.
They should be
P(7R3B|bag1) = 
P(7R3B|bag2) =  .
The use of Bayes' rule is correct.
|
|
|
