Can someone help settle a debate we are having in my team?
Four balls are placed in a bag. One is red, one is blue and the other two are yellow. The bag is shaken and someone draws two balls from the hat. He looks at the two balls and announces that at least one of them is yellow. What are the chances that the other ball he has drawn out is also yellow?
At first glance you may say 1/3, as you hold a yellow ball so 1/3 chance one of the remaining three balls will be yellow.
However I think the answer is 1/5 due to the independent and simultaneous drawing of the balls.
Consider the drawing possibilities:
YY
YB
BY
YR
RY
BR
RB
Knowing we hold one yellow ball there is no chance BR/RB occurred so the odds are left at 1/5.
Is this correct?
And if so can someone prove this mathematically?
Is this a probability paradox? Looking back after the balls are taken you can remove the BR/RB but does this invalidate the question?
Best Answer
"At first glance you may say 1/3, as you hold a yellow ball so 1/3 chance one of the remaining three balls will be yellow."
Well, that assumes you have a specific ball in mind to be "the" yellow that is yellow and a specific ball in mind to be the "other" ball. But you don't. Either ball (or both) can be "the" yellow ball and either (or both) can be the "other".
Think of it this way: If you say the ball in his right hand is yellow then the probability that the ball in his left hand is yellow is 1/3. And if you say the ball in his left hand is yellow then the probability that the ball in right hand is yellow is 1/3. But you are asking if either the ball in his left hand or the ball in his right is yellow what is the probability that the balls in both hands are yellow.
Those are obviously different questions.
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Actually your list should have 12, not 7, options.
$Y_1Y_2, Y_1B, Y_1R, Y_2Y_1, Y_2B, Y_2R, BY_1, BY_2, BR, RY_1,RY_2, RB$.
And of the $10$ remaining there are $2$ options.