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In a box, I have a ball, which has an unknown colour (can be either red or blue). I put in another ball, which is red, shake the box and take out a random ball. The ball taken out turns out to be red. If I take out the another ball, what is the chance of drawing a red ball?
Right after I read that question, I'm thinking the answer is 50%. The problem is, I have to have some sort of written proof. Can someone help me with that?
Hello, Grey!
The answer of 50% is much too easy.
. . It is a classic trick question.
There are four equally likely outcomes.Quote:
In a box, I have a ball, which has an unknown colour (can be either red or blue).
I put in another ball, which is red, shake the box and take out a random ball.
The ball taken out turns out to be red.
If I take out the another ball, what is the chance of drawing a red ball?
(A) The box contains a red ball, call it R1.
. . .A red ball is added, call it R2.
. . .(1) We draw R1, then we draw R2.
. . .(2) We draw R2, then we draw R1.
(B) The box contains a blue ball, call it B1.
. . .A red ball is added, call it R2.
. . .(3) We draw R2, then we draw B1.
. . .(4) We draw B1, then we draw R2.
We are told that the first ball is red.
. . Hence, case (4) is discarded.
We have three equally likely outcomes:
. . (1) We draw R1, then we draw R2.
. . (2) We draw R2, then we draw R1.
. . (3) We draw R2, then we draw B1.
In two of the three outcomes, the second ball is red.
Therefore: .P(second ball is red) .= .2/3
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This is identical to another classic.
I know my neighbor has two children.
I meet one of them and the child is a girl.
What is the probability that the other child is a girl?
The "obvious" answer is 50%; the child must be a boy or a girl, right?
. . Wrong!
There are three possible cases.
He has one girl, one boy.
. . .(1) You met the girl; the other is a boy.
He has two girls: G1 and G2.
. . (2) You met G1; the other is G2.
. . (3) You met G2; the other is G1.
In two of the three cases, the other child is a girl.
Oh, I should have seen that. My mind went instantly to 50% when I first read the question. Thanks