PrajayM wrote:
IanStewart The way I saw it is that if we are picking cards one after another that is the same as without replacement. So if we want a jack on the second pick that answer would be 7/17 (denominator reflecting the one card which is already picked and unavailable). So there would be two situations. The first card being a jack and second being a jack. Or the first being not a jack and the second being a jack. Where am I making a mistake in my thinking. Thank you in advance
That's a very common question, and you can certainly solve by breaking the problem into two cases. You'll get the same answer as I did; when the first card is a Jack, the probability the second is a Jack goes down, but when the first card is not a Jack, the probability the second is a Jack goes up. When you do the calculation, you'll see that here, those two situations balance out.
You can see conceptually why the answer should still be 7/18 here, even though we're picking the second card rather than the first. Here you can imagine removing the first card from the top of the deck. Then just put that card on the bottom of the deck. Now the card we're interested in is the first card in the deck, and we have the same deck as before - a deck of 18 cards, where 7 of the cards are Jacks - and we're picking one card from the top. So the probability will be 7/18 that it is a Jack.
In general, when you're picking without replacement, and you're interested in the probability your second (or fifth, or tenth, etc) selection is of a certain type, then if you have no information about your earlier selections, then it doesn't matter that the question asks about the second (or fifth, or tenth) selection. You can just pretend it's the first selection. But if you do have some information about the earlier selections, you must account for that when answering the question. So here, if you knew the first selection was a Jack, for example, then the answer is no longer 7/18 (instead it would be 6/17).
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