Question

We have two bags containing marbles: • Bag A contains 4 blue, and 3 green marbles • Bag B contains 2 blue, 5 green, and 3 yellow marbles We perform this two step procedure. First, draw a marble from bag A uniformly at random. Then put this marble into bag B. Second, draw a marble from bag B uniformly at random. We keep this marble. What is the probability that the marble we keep is green?

Answer 1

4B 28 SG 3Y A B and time เว Ewo Ways we get Grees naibles & put it in B and & Take green nouble from a groen nauble from take OR Take non- goes maible Put it B 4 (ie blue) from A and grouo nauble from B take PC getting group from B) = P( gras from A x P (gperman rows from A + Prblue from A x Pl grown fool blue from A) + x 18 20 77 77 = 38 ㅋㅋ 04935

A () Blue Green R3 SDG 31 3 QD미 5G 3y 먀 38 11

Answer 2

SOLUTION  :

 P(green from bag A)

= Green marbles in the bag / Total marbles in the bag

= 3 / (4 + 3)

= 3/7

P(blue from bag A)

= blue marbles in the bag / Total marbles in the bag

= 4 / (4 + 3)

= 4/7

The marble chosen from the bag A is added to the bag B.

So,

P(green from bag B after addition of marble of chosen from bag A)

= P(green bag A) * P(green bag B | (bag A green) + P(blue bag A) * P(green bag B | bag A blue)

= 3/7 * ((5 + 1) / (2 + (5 + 1) + 3)  + 4/7 * 5 / ((2 + 1) + 5 + 3)

= 3/7 * 6/11 + 4/7  * 5/11

= 38/77

= 0.4935 (ANSWER).