Given information:
- We have one box (B1) with 2 blue balls and 3 red balls.
- We have one box (B2) with 2 blue balls and 1 red ball.
- Next, two balls from B1 are placed into B2 (randomly selected).
- Then two balls are randomly drawn from B2, and it turns out to be two balls of different colours.
What is the conditional probability that two balls of the same colours were transferred from B1 to B2?
What I have tried so far:
- ways to pick two balls from B1: 5C2 = 10
- ways to pick two balls from B2: (3+2)C2 = 10 (again)
But then I don't know what the next steps should be, can someone help me?
[20:19] Edit:
I think that I made a mistake with the possible combinations, since it does not matter if we take for example BlueBall1 first and then BlueBall2. The possibilities are in that case: [{blue, blue}, {blue, red}, {red, red}]
The formula for conditional probability is known for me, but I don't understand how to apply it in this case.
Best Answer
lets model out how transfer from b1 to b2 can take place :
case 1) b1 transfers two blue balls to b2 , prob = 0.1
case 2) b1 transfers two red balls to b2 ,prob = 0.3
case 3) b1 transfers a red and a blue ball to b2 , prob = 0.6
lets model the picking of balls from b2 :
case 1.1) picking different colour balls from b2(4b,1r) after case 1 : prob = 0.4 * 0.1 = 0.04
case 1.2) picking same colour balls from b2(4b,1r) after case 1 : prob = 0.6 * 0.1 = 0.06
case 2.1) picking different colour balls from b2(2b,3r) after case 2 : prob = 0.6 * 0.3 = 0.18
case 2.2) picking same colour balls from b2(2b,3r) after case 2 : prob = 0.4 * 0.3 = 0.12
case 3.1) picking different colour balls from b2(3b,2r) after case 1 : prob = 0.6 * 0.6 = 0.36
case 3.2) picking same colour balls from b2(3b,2r) after case 1 : prob = 0.4 * 0.6 = 0.24
given that the balls selected from b2 are different , cases 1.1 , 2.1 , 3.1 apply . Out of these cases case 1 and 2 had same color transfer from b1 to b2 , case 3 had different colour transfer from b1 to b2 :
prob ( same color transfer from b1 to b2 | differnt colour selection from b2 )
= ( case 1.1 + case 2.1 ) / (case 1.1 + case 2.1 + case 3.1 )
= (0.04+0.18)/(0.04+0.18+0.36) = 0.38
or if you prefer formulas :
$$P(\frac{same:b1->b2}{diff:b2})*p(diff:b2) = P(\frac{diff:b2}{same:b1->b2})*P(same:b1->b2)$$
$$P(\frac{same:b1->b2}{diff:b2}) = \frac{P(\frac{diff:b2}{same:b1->b2})*P(same:b1->b2)}{P(diff:b2)}$$
$$P(\frac{diff:b2}{same:b1->b2})*P(same:b1->b2) = case 1.1 + case 1.2$$ $$P(diff:b2) = case 1.1 + case 2.1 + case 3.1$$ $$P(\frac{same:b1->b2}{diff:b2}) = \frac{case 1.1 + case 1.2}{case 1.1 + case 2.1 + case 3.1}$$ $$P(\frac{same:b1->b2}{diff:b2}) = \frac{(0.04+0.18)}{(0.04+0.18+0.36)} = 0.38$$