Another Cards Problem : GMAT Quantitative Section
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# Another Cards Problem

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Joined: 16 May 2012
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Another Cards Problem [#permalink]

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25 May 2012, 12:19
I take advantage of this first post to say hi to the GMAT Club which I find very interesting and useful.

I experience some difficulties with a simple probability problem found in the book "Conquering GMAT Math and Integrated Reasoning":

"What is the probability that a card selected at random from a standard deck of playing cards will be a 7 or a club ?"

It's a example to illustrate some theory so there is no multiple choice but the answer is:
"There are 52 cards in a deck. There are 13 each in hearts, clubs, spades, and diamonds. There are four 7s, one each in hearths, clubs, spades, and diamonds. Thus the 7 of clubs is counted twice , once as a club and once as a 7 There are 13 clubs plus tree 7s that are not clubs.
P(7 or Club) = 16/52 = 4/13

My problem with this answer is actually the fact that this 7 of clubs isn't counted twice as announced : why doesn't this card count as a double chance of getting an expected card (7 or club) ?
My natural answer would have been P= (13+4)/52 = 17/52

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Re: Another Cards Problem [#permalink]

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25 May 2012, 13:07
I honestly don't know a great explanation for you other than use logic. You should not and cannot double something in a probability...at least in this setup. Just because a card meets multiple criteria it does not mean it needs to be counted multiple times.

The trick that they want you to catch on to is the fact that you CANNOT count the 7 of clubs twice.
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Re: Another Cards Problem [#permalink]

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25 May 2012, 14:02
I see.

GMAT study make your mind looping sometimes.

At the end we consider the probability of the instance itself (the 7 of club) and not it's value (7 and clubs).

Thank you for answering.
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Re: Another Cards Problem [#permalink]

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27 May 2012, 05:06
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Hi sinclairv,

It is by definition that Probability of an event = (number of favorable cases)/(exhaustive number of cases)

The number of favorable cases would be 16, 'cause even if 7 of club is picked - although it fulfills the criteria of being a club or being 7, it will be counted as 1 card only.

And therefore the probability is 16/52.

So, whenever dealing with probability questions, try to visualize a real life example...and it will make the problem easy.

Regards,
Re: Another Cards Problem   [#permalink] 27 May 2012, 05:06
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# Another Cards Problem

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