You face an urn with 5555 cards in it, each has a non-zero integer written on it. Nothing is said about the distribution of those numbers. You are told to draw randomly a card, copy the number, return it back, shuffle and draw randomly a card, then write down the sum of both numbers, say S.
(i) Prove: The probability of S being an even number is higher than S being odd.
(ii) Is it true for any initial number of cards? Comment.
(In reply to
Argument (no math) by Steve Herman)
Even if the initial number were not odd, say 4444, the probability that the distribution would be exactly evenly divided between odd and even would be tiny (for such large numbers). But even for small numbers, an even sum is more likely:
1 card: each draw is the same, so the total is twice that value, obviously even.
2 cards: The probability is 1/2 that the two cards have the same parity, assuring an even sum; and 1/2 they are of opposite parity, making an even sum 50% likely. The overall probability of an even sum is therefore 1/2 + 1/2 * 1/2 = 3/4.
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Posted by Charlie
on 2017-02-14 10:03:01 |
re: Argument (no math)
