I have a black box that contains 4 balls. 3 are red and 1 is blue.
I remove two from the box without seeing them.
I look at one of them, it is red.
What is the probability that the other one is red?
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The procedure of drawing 2 balls and looking at 1 can produce 12 equally likely outcomes. We can draw balls in 6 different ways, and we can choose either of the 2 balls we drew to look at. Labeling the red balls R1, R2, and R3, and using an arrow to mark the one we look at, the possibilities are
R1,R2 R1,R3 R2,R3 R1,B R2,B R3,B ^ ^ ^ ^ ^ ^ R1,R2 R1,R3 R2,R3 R1,B R2,B R3,B ^ ^ ^ ^ ^ ^
When we see a red ball, this evidence eliminates 3 possibilities:
R1,B R2,B R3,B ^ ^ ^
Of the 9 possibilities consistent with the evidence, 6 of them have the other ball be red. Thus, the probability that the other ball is red
It's equivalent to taking one of the balls, then looking at it, then taking out another and looking at it. 66% is my guess.
Edit: I am going to risk ending up stupid and complain: why 50%? Maybe the combinatorics data ends up showing that but to me it doesn't make any sense. I'm going to stick with 66% due to the fact that, if we assume we have balls labelled
R1 R2 R3 B, giving six possible matches, when we pick out two balls and look at one, we can determine that one of the balls is, for example,
R1. This means that there are three valid matches:
R1R2 R1R3 R1B and three invalid ones:
R2R3 R2B R3B.
R1R3 thus make up two thirds of the possible matches.
If I am thoroughly wrong in my methodology, might someone explain why? The other answers I've seen don't really cut it for me but that might just end up being chalked up to pickiness or stubbornness.
This is how I see it Basically you have 4 balls. What is the probability that you will pick 2 red balls out of the 4?
There are only 6 ways to choose 2 balls out of the 4. And only 3 of those will give you both red balls.
So my answer is:
3/6 = 1/2
Since @w0lf posted the clarification I am going to try to do this one with the proper math notation.
Let A denote that first ball is red.
Let B denote that the second ball is red.
Then P(B|A) = P(A and B)/P(A)
Read P(B|A) as probability of B happening after A has happened.
P(A and B) i have solved above. which is equal to 1/2. And P(A) is 3/4. Therefore P(B|A) = (1/2)/(3/4) = 2/3
Classical conditional probability (or "Bayesian" probability) problem. Bayes formula is well known, but just for simplicity purposes,one may simply think: I have one ball FIRMLY in hand and it IS red. So the problem has transformed to: Draw one ball out of 3 (2 red, 1 blue) probability of drawing a red, clearly= 2/3
To expand on DarkGammas answer
Label the balls R1, R2, R3 and B1 if you grab two of them the possible combinations are
R1/R2 R1/R3 R1/B1
R2/R1 R2/R3 R2/B1
R3/R1 R3/R2 R3/B1
B1/R1 B1/R2 B1/R3
For a total of 12 combinations.
You take two out and look at one which happens to be red. We will call this one R1.
Now if you know the one you looked at was the first ball grabbed you can narrow it down to:
R1/R2 R1/R3 R1/B1 (3 combinations; 2 combinations have the second ball being red; 2/3)
Now if you know the one you looked at was the second ball grabbed you can narrow it down to:
R2/R1 R3/R1 B1/R1 (3 combinations; 2 combinations have the first ball being red; 2/3)
If you do no know which one you are looking at your possible combinations are:
R1/R2 R1/R3 R1/B1 B1/R1 R2/R1 R3/R1. (6 combinations; 4 combinations have both red;2/3)
Therefore I believe 66.6% to be the answer.
(I was initially working under the premise that the last example would yield a result of 50%. once I found that to be wrong, I didn't want to waste the time I put into this answer and posted it anyways.)