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This is a problem of conditional probability:

There are two boxes X and Y. X contains 5 balls: 2 are blue, 2 white and 1 is gray. Y contains 4 balls: 2 are blue, 1 is white and 1 is gray.

I have to calculate the probability of both balls being white given that both are of the same colour. One ball is going to be taken from each box.

This problem, at least when I look to the formula of conditional probability ((P(A|B) = P(A and B) / P(B)), isn't very intuitive to me.

In this case, shouldn't the probability be P(White and White | Same Colour) = P[(White and White) and Same Colour)/P(Same Colour)?

My text book says it's going to be P(White and White) / P(Same Colour).

Could you guys help? Thank you very much and sorry about the wall text.

Edit: missed a sentence.

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    $\begingroup$ I think you're missing a sentence. Is one ball drawn from each box? $\endgroup$ Nov 2 '17 at 13:01
  • $\begingroup$ Yes, that is relevant information that I have forgotten. I am going to edit that, thank you. $\endgroup$ Nov 2 '17 at 13:13
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You are both right. What you haven't yet noticed is that if the balls are both white, then they necessarily have the same colour. So we have $$ P((\text{white and white})\text{ and same colour}) = P(\text{white and white}) $$ Thus you have used the formula directly, while your text book has used the formula then applied a small simplification. Note that this simplification, while small, is a relatively crucial step towards being able to calculate anything.

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  • $\begingroup$ Oh, I get it now. This was illuminating, Arthur. Thank you so much for your answer. $\endgroup$ Nov 2 '17 at 17:48
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The general formula that you point to does of course work.

But in this case there is another way to think about it, which is simply 'Out of how many possible pairs of balls that have the same color, how many are a white pair?' And with that line of thinking you get the book's formula... but the same answer.

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  • $\begingroup$ This made me a little bit confused. Looking through what you said, "out of how many pairs of balls that have the same color, how many are a white pair?", shouldn't this be P(x) = 1/3? There are three possible pairs with the same color and one of them is a white pair. Where is my mistake? Thank you for your answer. $\endgroup$ Nov 2 '17 at 17:52
  • $\begingroup$ @BrunoNascimento Because there are 3 colors? No, clearly it doesn;lt work that way. Suppose you just pick one ball from box X ... would you say that the probability of getting a white ball is 1/3 because there are 3 colors? No, you wouldn't (I hope!), because the number oif instances oif each color effects that probability. Well, same with the number of possible pairs: for example, because you have 2 blue balls in box X, and 2 blue balls in box Y, there are 4 ways to get two blue balls ... but there is only 1 way to get a pair of white balls. $\endgroup$
    – Bram28
    Nov 2 '17 at 19:38

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