the odds theoretically are almost infinity to 1 as they could be any color. -------------------------------------------------------------------------------------------
2nd opinion
Let's say we have 3 boxes with 4 balls each.
Box A has 4 white balls.
Box B has 3 white balls.
Box C has 2 white balls.
The probability of drawing 2 W balls from;
Box A, P(2W│A)=
(4/4)∙(3/3)=
1
Box B, P(2W│B)=
(3/4)∙(2/3)=
1/2
Box C, P(2W│C)=
(2/4)∙(1/3)=
1/6
Say the probability of picking any of the 3 boxes is the same, we have;
P(A)=
1/3
P(B)=
1/3
P(C)=
1/3
Question is, given the event of drawing 2 W balls from a box taken blindly
from the 3 choices, what is the probability that the balls came from box A,
P(A│2W).
Recurring to Bayes Theorem:
P(A│2W)=
[P(A)P(2W│A)]/[P(A)P(2W│A)+P(B)P(2W│B)+P(C)P(2W│C)]=
[(1/3)(1)]/[(1/3)(1)+(1/3)(1/2)+(1/3)(1/6)]=
6/10=
0.60=
60%
P(A│2W)=
0.60=
60%
1/4 chance
The probability is (6/14)*(5/13) = 30/182 = 0.1648 approx.
(3/7)*(2/7)=(6/49) You have a 6 out of 49 probability.
If you draw enough balls, without replacement, the probability is 1.The answer depends onhow many balls are drawn, andwhether or not they are replaced.Unfortunately, your question gives no information on these matters.
3c2+2c2+3c1 2c1 --------------- 7c2. I have done it this way. 3C2 means that the both balls would be green 2C2 both would be red and 3C1 and 2C1 because both the balls can be red and green.
The probability that it contains exactly 3 balls is 6/45 = 0.133... recurring.
If the balls are selected at random, then the probability is 12/95.
40/50
1/4 chance
It is 1/7 if a ball is selected at random.
17 out of 21
12
The probability is (6/14)*(5/13) = 30/182 = 0.1648 approx.
(3/7)*(2/7)=(6/49) You have a 6 out of 49 probability.
2
It is 1/6.
If you draw enough balls, without replacement, the probability is 1.The answer depends onhow many balls are drawn, andwhether or not they are replaced.Unfortunately, your question gives no information on these matters.