There are three bags in a locker room. Bag 1 has three white balls and two black

Question

There are three bags in a locker room. Bag 1 has three white balls and two black balls. Bag 2 has four white balls and three red balls. Bag 3 has two white balls and three red balls. Your first friend selects a bag at random and draws a ball (no replacement). Your second friend selects a bag at random and draws another ball (no replacement). Both balls are white. What is the probability that both balls were drawn from bag 1?



There are three bags in a locker room. Bag 1 has three white balls and two black balls. Bag 2 has four white balls and three red balls. Bag 3 has two white balls and three red balls. Your first friend selects a bag at random and draws a ball (no replacement). Your second friend selects a bag at random and draws another ball (no replacement). Both balls are white. What is the probability that both balls were drawn from bag 1?

Explanation / Answer

Bag 1 - 3W, 2B

Bag 2 - 4W, 3R

Bag 3 - 2W, 3R

Let event A -> Both balls drawn are white

Let event B -> Both balls are drawn from Bag 1

We need to P(B/A) = P(both balls are from Bag 1 given that both balls drawn are white)

From Bayes' Theorem,

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

P(A/B) = P(both balls are white given that both balls are from bag 1) = 3/5 x 2/4 = 3/10

P(A/B) = 3/10 = 0.3

P(B) = P(both balls are from bag 1) = P(first friend selects bag 1 and second friend selects bag 2) = P(first friend selects bag 1) x P(second friend selects bag 2) = 1/3 x 1/3 = 1/9

P(B) = 1/9

By Total Probability Theorem,

P(A) = P(A/B)P(B) + P(A/B')P(B')

We now need to define B' : If B = both balls are from bag 1 --> (1,1)

then B' = both balls are NOT from bag 1 --> (1,2) (2,1) (2,2) (1,3) (3,1) (3,3) (2,3) (3,2)

And P(A/B') = P(both balls are white given that both balls are NOT from bag 1)

Let Bij = Balls are drawn from Bag i and then Bag j

For eg : B12 = Balls are drawn from Bag 1 and then Bag 2

----> P(A/B')P(B') = P(A/B12)P(B12) + P(A/B21)P(B21) + P(A/B22)P(B22) + P(A/B23)P(B23) + P(A/B32)P(B32) + P(A/B33)P(B33) + P(A/B13)P(B13) + P(A/B31)P(B31)

Note that P(B12) = P(B21) = .... = P(B33) = 1/9 (As each of them makes up 1 outcome out of 9 possible ways of picking 2 bags out 3 bags)

----> P(A/B')P(B') = (3/5)(4/7) + (4/7)(3/5) + (4/7)(3/6) + (4/7)(2/5) + (2/5)(4/7) + (2/5)(1/4) + (3/5)(2/5) + (2/5)(3/5) / 9

----> P(A/B')P(B') = 12/35 + 12/35 + 12/42 + 8/35 + 8/35 + 2/10 + 6/25 + 6/25 / 9

----> P(A/B')P(B') = 360 + 360 + 300 + 240 + 240 + 210 + 252 + 252 / 1050 x 9

----> P(A/B')P(B') =  0.234

P(A) = P(A/B)P(B) + P(A/B')P(B')

P(A) = (3/10)(1/9) + 246/1050

P(A) = 1/30 + 246/1050 = 35+246/1050 = 281/1050

P(A) = 281/1050

Final formula ---->

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

P(B/A) = (3/10)(1/9) / 281/1050

P(B/A) = 1/30 / 281/1050 = 0.1245

Answer : P(B/A) = P(both balls are from Bag 1 given that both balls drawn are white) = 0.1245

Cheers!

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