Three bags of marbles
You have three bags of marbles. The first bag contains two white marbles, the second contains two black marbles and the third contains one black and one white marble.
You pick a bag at random and take out a marble. It is white. What is the probability that the remaining marble in that bag is white?
You pick a bag at random and take out a marble. It is white. What is the probability that the remaining marble in that bag is white?
Labels: Probability
posted by Chris at
1:09 PM
8 Comments:
it would be 2/6 or 1/3 simplifyed
At least you didn't say 1/2. But you also didn't say the correct answer.
P(A/B)=P(A and B)/P(B)
A = 2nd white
B= 1st white
P(A and B)=1/3 only one bag allows this
P(B)=1/3*1/2+1/3*1+1/3*0=1/6+1/3=1/2
P(A/B)=P(A and B)/P(B)=(1/3)/(1/2)
P(A/B)=2/3
or to think of it another way
if I have picked 1 of 3 white marbles, I have chosen one of the 2 in the double white bag (2/3) or the 1 in white/black bag (1/3).
the 2/3*100%+1/3*0%=2/3 is the chance the other marble in the bag is white
Cam
Needless to say, Cam got it right and for the right reasons.
ok...same question,but this time i have added 1 more white marble to each bag.
Knightmare. Your question is answerable. What is "THE remaining marble"?
It would be a 50/50 chance if the second is white.
This is because if the first marble you pull out is white then that means you pulled the bag that had either both white marbles.. or the one with a white and black marble. then making that between two bags. Becuase there is no way you pulled the bag with two black marbles if you pulled out a white one on the first draw. so the bag with two black marbles can not be a factor in this
Hi Alex. I think you meant it's 50/50 that the second marble is white. That's wrong, it's 2/3.
You are right in that the third bag is irrelevant. It was put into to add to the challenge.
Cam's given a couple of explanations.
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