find the probability that 2 green balls are selected from box A.?

Question

If 2 red balls are selected, what is the probability that they came from box B?

There are 3 red balls and 3 green balls in box A, and 4 red balls and 5 green balls in box B. An experiment consist of first choosing a box at random and then selecting from it 2 balls, one after the other without replacement.

Answer 1

CASE 1:
The probability that 2 green balls are selected from box A:
1/2 (chance of picking box A)
Chance of picking first ball green (3/6)
Chance of picking second ball green (2/5)

So the combined odds are: 1/2 x 3/6 x 2/5 = 1/4 x 2/5 = 1/10

CASE 2:
The probability that 2 red balls are selected from box B:
1/2 (chance of picking box B)
Chance of picking first ball red (4/9)
Chance of picking second ball red (3/8)

So the combined odds are: 1/2 x 4/9 x 3/8 = 12/72 = 1/6

Answer 2

The probability that box A is selected is 1/2.
The probability that the first ball is green is 3/6
The probability that the second ball is green is 2/5

So, the probability of 2 green balls from A is 1/2 * 3/6 * 2/5 = 1/2 * 1/2 * 2/5 = 1/10 = 0.1

As for the second question, you may need to be a little clearer about exactly what you mean but I would look at it this way.
The probability of 2 red balls from A is 1/10
The probability of 2 red balls from B is
1/2 * 4/9 * 3/8 = 1/12

So, if 2 red balls are selected, the probabilty tha they came from B is (1/12)/(1/10 + 1/12) = 0.4545.

Answer 3

No disagreement with Case 1.
Case 2. Probability of 2 red balls selected from Box A
1/2 Chance of picking A.
3/6 Chance of picking first ball red
2/5 Chance of picking second ball red
= 1/2 * 3/6 * 2/5
= 1/10

Probability of 2 red balls selected from Box B
1/2 Chance of picking B
4/9 Chance of picking first ball red
3/8 Chance of picking second ball red
= 1/2 * 4/9 * 3/8
= 1/12

Total chance of picking 2 red balls = 1/10 + 1/12 = 6/60 + 5/60 = 11/30.

So IF 2 red balls WERE picked, chance that it was box B:
[1/12] / [11/30]
= 5/60 / 11/30
= 5/11

Answer 4

There are 2 procedures to respond to this. a million. Assuming we replace decrease back the selected balls, then: P = 3 / (6 + 4 + 2) = 3 / 12 = a million / 4 = 0.25 = 25% 2. Assuming we do not replace decrease back the selected, then: P = a million/12 + a million/11 + a million/10 = 181 / 660 = 0.274 or 27.4% desire this allows.