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Probability with cards

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Probability with cards

by praveen_gmat » Wed Oct 12, 2011 7:28 am
When four cards are drawn in succession from a pack of cards with replacement, what is the probability that all the cards are from different suits?
Note: A suit is Diamonds(13), Hearts(13), Spades(13) or Clubs(13).


ANS: 13C1/52C1 * 13C1/52C1 * 13C1/52C1 * 13C1/52C1 = 1/256

But for me, it should have been:

13/52 * 13/39 * 13/26 * 13/13.
Because, the question asks for different suits each time, the second time I pick, i should not have any one 1st 13 cards(suit) and so on. ..

Any idea?
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by shankar.ashwin » Wed Oct 12, 2011 7:36 am
You have 4 suits in a pack of card (with equal number of cards each).

Prob of picking up any suit is 1/4 and prob of picking up a suit is independent of each other.

So you have equal chances of picking up any suit.

4 diff suites - (1/4)*(1/4)*(1/4)*(1/4) = 1/256
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by GmatMathPro » Wed Oct 12, 2011 8:31 am
On the first pick there is a 100% chance of getting SOME suit. On the next draw you have a 3/4 chance of getting a different suit. 3rd draw you have a 2/4 chance of getting something different from the first two. 4th draw you have a 1/4 chance of getting something different from the first 3 draws:
1*(3/4)*(2/4)*(1/4)=3/32

OR

using counting: To get 4 different suits it has to be some arrangement of HSDC. There are 4! arrangements of this, out of 4^4 possible sequences for suits. 4!/4^4=3/32
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by praveen_gmat » Wed Oct 12, 2011 9:22 am
@GmatMathPro:
The answer is 1/256
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by shankar.ashwin » Wed Oct 12, 2011 9:27 am
I think 1/256 is wrong. I saw the answer and tried to answer the question.

If the question was what you posted it should be 3/32 only. GmatMathPro is bang on with the solution
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by praveen_gmat » Wed Oct 12, 2011 9:38 am
Well, i have this question from a leading CAT tutoring institute's book. The answer is 1/256

By the way, the way I see it is...
13/52 * 13/39 * 13/26 * 13/13.
Every time i pick the card, the suit of the card should be different from what I picked earlier. So at first, i have 52 cards out of which the favorable outcomes are any 13. Next time, I have to pick any of the remaining suits ( 3 suits left). SO it has to be 13 out of 39.

While I am writing I feel I am wrong, but I am not convinced by any of the answers.
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by GmatMathPro » Wed Oct 12, 2011 9:53 am
I'm not sure I'd trust people who tutor cats for a living.

But anyway, let me ask you this. Why are the favorable outcomes only 13 out of 52 on your first card? That means that 39 of the 52 cards are unfavorable. If you picked any one of them, you'd say "oh crap, i guess i'm screwed for getting 4 different suits." But what are those 39 cards???
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by praveen_gmat » Wed Oct 12, 2011 10:02 am
yeah, what I said is wrong. I realize.
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by jsnipes » Wed Oct 12, 2011 10:19 am
so with replacement there will always be 52 cards and an equal number of suits. using the slot method you can pick any card in first slot, 52/52 * 39/52 (52 cards -13 cards of whichever suit was picked first) * 26/52 (52- 13*2 of the two suits picked so far) * 13/52 (only one suit remains)

or, 1*(3/4)*(1/2)*(1/4)=3/32 i believe is the correct answer.
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by GmatMathPro » Wed Oct 12, 2011 10:28 am
Also, notice that 1/256 is the probability of getting some specific sequence of 4 different suits. Like if you specifically wanted to get hearts, then diamonds, then spades, then clubs. It would be (1/4)^4=1/256. But if we want the probability of ANY sequence of different suits then the probability must increase.

It seems to be extremely difficult to publish a test prep book without any errors whatsoever. I think sometimes what happens is they decide to make slight modifications to the wording of a problem for whatever reason and then they forget to make the corresponding modification to the solution. You might try searching for a list of known errata for whatever book you're using.
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by praveen_gmat » Wed Oct 12, 2011 10:34 am
Agree !
Your method should help me solve other problems.
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