Quant Probability Question – make my head itch!

This means this is a “1 – probability()” question…this is the default strategy when a question is asking the probability of at least X happening. It translates into 1 – probability(that event NOT happening)

Step 2:
Now you need to find the probability of not getting a pair for each of the 4 cards. Recognize that as you go through each card, the number of available cards from the remaining deck is constantly changing. Originally you have 12 cards, but then after you put down 1 card, u have 11 cards left.

Step 3:
1st card = any card = prob of 1 or 100%
2nd card = any card except the value of first card = 10 out of 11 choices
3rd card = any card except first card and except second card = 8 out of 10 choices
(notice we must exclude the 1st card and the 2nd card–but also the corresponding pairs for each of these cards.
4th card = any card except first/second/third card (and their corresponding pair card) = 6 out of 9 choices

Now multiply these probabilities so you know what the probability of NOT getting any pairs for all 4 iterations.
10/11 * 8/10* 6/9 = 480 / 990

1 – 480/990 = 510/990 = 17/33

Answer choice is C (17/33)

I don’t recommend using the nCr formula here, but if you want to use nCr’s, then:

= 1 – prob(choosing all different cards
= 1 – prob(choosing 4 different cards out of 6) * (possibilities for first card to be of two suits, for second card to be of two suits, for third card to be of two suits, and for fourth card to be of two suits)
= 1 – (6c4 * 2 * 2 * 2 * 2) / (12c4)
= 1 – (15 * 16) / (( 12 * 11 * 10 * 9 )/(4*3*2*1))
= 1 – 240 / 495
= 255/495

= 51/ 99
= 17 / 33

But again, this is not the most efficient way to think through this question. The best way is to count according to the method I discussed first above.
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Zeke Lee
Co-Founder, The GMATPill Study Method

http://www.gmatpill.com

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