Isn't it just W/(W+B) ... What is this ,.,GCSE ...teacher couldn't do this?!
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Posted from TSR Mobile
(edited 10 years ago)
Original post by #Unknown
Hey, could someone help me out with this maths problem. My maths teacher couldn't solve it so I didn't where to look for the answer. The question is 'If I have B Black balls and W White balls, what is the probability that the ball I select is white?' Thanks! 
This belongs in the Maths study help forum - if you're going to reply to this post, make a new post in that forum, to avoid mixing this one up
There are B+W possible choices of ball I could make. Out of those B+W choices, W of them are white. Hence W/(B+W) is the probability I pick a white ball, because I'm equally likely to pick any ball. (Intuitively, this makes sense - you can carry out the experiment for some small B and W.)
Original post by #Unknown
Sorry, the last bit was meant to say 'what is the probability that the LAST ball I select is white'. Meaning that you take out the balls from a bag and what's the probability that the last ball I take out is white.
Makes no difference; it's still W/(W+B) - assuming they're being taken out at random.
(edited 10 years ago)
Original post by #Unknown
It can't be, thats the probability of taking a white ball first, then the probability of picking a white ball second is W-1/W+B-1 because there are 1 fewer balls in the bag.
The probability of picking a white ball second is still W/(B+W) even though there is one fewer ball in the bag. The expression you have written assumes that you know that the first ball which was removed was a white one rather than a black one.
If the first ball removed is white, then:
There are now W-1 White balls and B Black balls, so the probability of the next one being white is (W-1)/(W-1+B)
If the first ball removed is black, then:
There are now W White balls and B-1 Black balls, so the probability of the next one being white is (W)/(W+B-1)
What if you don't know the colour of the first ball that was removed?
We want to know the probability that the second ball is white. There are two possibilities: Either the first is white and the second is white, or the first is black and the second is white. The probabilities of these, respectively, are:
(W)/(W+B) x (W-1)/(W-1+B)
and
(B)/(W+B) x (W)/(W+B-1)
If you add these together and simplify, you end up with just (W)/(W+B). A similar thing will happen if you're talking about the third ball, fourth ball or the last ball.
(edited 10 years ago)
Original post by #Unknown
wow! I can't believe it's that simple. I asked a different maths teacher and he came up with 1/W as the answer.
Since the balls are picked at random, the probability of a white ball being left, is the same as the probability of choosing a white ball in the first place (or in any place), since they are essentially the same action: Choosing one ball at random.
Original post by #Unknown
wow! I can't believe it's that simple. I asked a different maths teacher and he came up with 1/W as the answer.
That would imply that the probability is unchanged however many black balls are there. That's clearly rubbish - imagine a universe full of black balls, and W white balls where W < 100.
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