Probability of drawing odd number of balls from an urn

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.

• If you draw an odd number of blues, then you draw an odd number of blacks and vice versa: these events are the same. Aug 27, 2020 at 1:28
• 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. Aug 27, 2020 at 2:12

The probability that they are both odd is $$1/2,$$ as you have shown, and the probability that they are both even is also $$1/2.$$
• 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 $0.5$... Aug 27, 2020 at 2:44