Probablity

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Probablity

by parveen110 » Sun Apr 27, 2014 5:03 am
Two cards are drawn at random from a pack of cards. What is the probablity that both these cards are of black color or both are aces?
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by [email protected] » Sun Apr 27, 2014 1:24 pm
Hi parveen110,

This type of probability question comes with a minor twist to the math: you have 2 black aces, so you have be careful that you don't count those "combos" more than once.

In real basic terms, probability is "the number of ways you want something to happen" divided by the "total number of possibilities."

Since we're dealing with (what I assume is) a 52 card deck, there are a couple of ways to do the "math"....

Assuming order DOES matter, the total number of possible ways to pull two cards is: 52 x 51

The total number of ways of pulling 2 black cards is: 26 x 25
The total number of ways of pulling 2 aces is: 4 x 3

BUT since two aces are black, that "combo" appears in both the "2 black cards" and the "2 aces" calculations. You're NOT allowed to count that combo twice, so you have to subtract the extra iterations.

The total number of ways of pulling 2 black aces is: 2 x 1

So, the final probability is:

[(26)(25) + (4)(3) - (2)(1)] / (52)(51)

This reduces to: 160/ 663

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by Brent@GMATPrepNow » Sun Apr 27, 2014 2:13 pm
Aside: this question, in its current form, could never be an official GMAT question, because it's culturally biased. The GMAT would never require test-takers to know the composition of a deck of cards, because not all cultures play cards.

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by Matt@VeritasPrep » Tue Apr 29, 2014 2:26 pm
Brent@GMATPrepNow wrote:Not all cultures play cards.
Say it ain't so!

(On a more serious note, if this question were on the GMAT, the testwriters would probably just tell you the composition of a deck of cards. They don't seem to mind assuming all cultures roll six-sided dice, however!)

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by Matt@VeritasPrep » Tue Apr 29, 2014 2:39 pm
Also, maybe I'm missing something, but I'm getting a different answer from Rich. (If I do it his way, the actual simplification I get is 166/663, but that doesn't seem to be right either.)

Assuming this is without replacement (i.e. both cards are drawn at once):

Prob(Black,Black) = (26/52) * (25/51) = 25/102
Prob(Ace,Ace) = (4/52) * (3/51) = 1/221

But we've overcounted the pair of black aces, as they appear in each of the above. So we subtract

Prob(Black Ace,Black Ace) = (1/26) * (1/51)

This gives us 25/102 + 1/221 - 1/1326 = 55/221. (Close, but just a hair off.)

Where I differ from the original post is the notion that there are 12 ways to pull four aces or 2 ways to pull two black aces. That assumes the order matters, but I'm not sure it does here: there's only one pair of black aces (Ace of ♠, Ace of ♣), for instance. Likewise, the # of pairs of black cards is (26 choose 2), so the probability of two black cards is (26 choose 2)/(52 choose 2).

If this is done WITH replacement, then we have

Prob(Black,Black) = (26/52)(26/52) = 1/4
Prob(Ace,Ace) = (4/52)(4/52) = 1/169
Prob(Black Ace, Black Ace) = (2/52)(2/52) = 1/676

So 1/4 + 1/169 - 1/676 = 43/169.

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by [email protected] » Tue Apr 29, 2014 9:17 pm
Hi parveen110,

Matt did catch my calculation error (the formula is correct, but I accidentally subtracted 10 instead of adding 10 to the numerator).

So, [(26)(25) + (4)(3) - (2)(1)] /(52)(51) =

660/2652 =

55/221

This is the same result that you'll get if you treat this prompt as a Combination question (which is how Matt approached it). Both approaches are based on the assumption that you're drawing two distinct cards from the deck (and not drawing the same card twice).

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