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There's an urn with equal number of blue and black balls. You draw 10 balls, with replacement (how would the question be different without replacement, by the way?). What's the probability of drawing an odd number of blue balls?

Here's my logic:

P(getting an odd number of blue) = (10C1 + 10C3 + 10C5 + 10C7 + 10C9)/1024 = 1/2

The P(getting an odd number of black) is also 1/2. How can this be possible, as there should still be some probability remaining for obtaining an even number of black/blue balls, however 1/2 + 1/2 = 1.

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    $\begingroup$ If you draw an odd number of blues, then you draw an odd number of blacks and vice versa: these events are the same. $\endgroup$ Aug 27, 2020 at 1:28
  • $\begingroup$ If there were just five blue and five black balls in the urn, then when you draw $10$ of them without replacement, you will ALWAYS get an odd number of blue balls, and that odd number will always be five. If there were six blue and six black and you draw $10$ without replacement, then you will always get at least four blue balls and never more than six, whereas if it's without replacement you could get $0$ blue balls or $10$ or anything between those. $\endgroup$ Aug 27, 2020 at 2:12

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Getting an odd number of blue balls and getting an odd number of black balls are not complementary events. In fact they are the same event: If you have 10 balls and each is either blue or black, then the number of blue balls and the number of black balls are either both even or both odd.

The probability that they are both odd is $1/2,$ as you have shown, and the probability that they are both even is also $1/2.$

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  • $\begingroup$ Interesting (to me at least): there are actually 11 final possibilities (R...RR, R...RB, ..., RB....B, BB...B). Among these 11 final configurations $5$ are odd and $6$ are even... however the distribution of these configurations make them not equiprobable. And the final answer is still $0.5$... $\endgroup$
    – Déjà vu
    Aug 27, 2020 at 2:44
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    $\begingroup$ @e2-e4 : Correct. Look at it this way: Take any particular sequence of blues and blacks, and look at the color of the first ball you draw. It's either blue or black. If it's blue, change it to black and vice-versa. Those two outcomes---where the first one is blue and where it's black---are equally probable. But one of them involves an odd number of each color and the other even. $\endgroup$ Aug 27, 2020 at 4:07

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