What is the probability that you get a pair when picking the top two c

Question

What is the probability that you get a pair when picking the top two cards off of a randomly shuffled deck of cards (52 total cards, with 13 sets of four matching suits)?

(A)  \frac{12}{2,652} 
(B)  \frac{16}{2,652} 
(C)  \frac{1}{17} 
(D)  \frac{1}{13} 
(E)  \frac{1}{2}

Pair of same design(set of 13) or matching suits(types of 4)?
Both give same answer.

Kurtosis · Answer

First pick --> Pick any card
Second pick --> Out of the remaining 51 cards, only 3 cards can match the first card

Probability = 3/51 = 1/17

Answer: C

bethebest · Answer

Did this -

52C1*3C1 /52C2

It is coming out to 2/17. What am I doing wrong here?
Thanks.!

BrentGMATPrepNow · Answer

P(select pair) = P(1st card is ANY card   AND   2nd card matches 1st card)
= P(1st card is ANY card)   x   P(2nd card matches 1st card)
= 1   x   3/51
= 3/51
= 1/17
= C

Aside: P(2nd card matches 1st card) = 3/51, because once 1 card is selected, there are 51 cards remaining in the deck. Among those 51 remaining cards, there are 3 that match the 1st card selected.

Cheers,
Brent

jps245 · Answer

hello,

I don't think you can use this distribution in this way because the numerator has overlapping sets. specifically, the numerator above says that you're choosing one card out of the 3 cards, and one card out of 52 cards which would includes those other 3 (in other words you don't want these sets to overlap in this type of distribution)

the best way of approaching this problem was mentioned above - no matter what you pick as the first card, there is a 3 / 51 chance of getting a pair (3 cards that will give you a pair out of the remaining 51)

hope this helps

longhaul123 · Answer

can someone please let me know where am I going wrong. 
Now for the first card the probability is 1/52
now there are 3 pairs left in the other 3 sets and there is a total of 51 cards left.
So the probability is 1/52*3/51
=1/52*1/17
why is this method incorrect ??

stne · Answer

Hi  longhaul123 ,
Hope I am not too late here , anyway.

For the first card the probability is  \frac{52}{52} and NOT  \frac {1}{52}  , because you can select any of the 52 cards out of a total 52 cards.
For the 2nd card we have only 3 options , because there are only 4 of the same number, four 1's , four 2's , .....four kings, four queens and four aces .So for any first number we pick there are only 3 other remaining of the same number.
Hence the probability is  \frac {52}{52} * \frac{3}{51} = \frac{1}{17}

Hope this helps. Let me know if anything is still unclear.

ScottTargetTestPrep · Answer

Suppose the first card has been chosen. Now, among the remaining 51 cards, there are 3 cards that form a pair with the already selected card and thus the probability of getting a pair is 3/51 = 1/17.

Answer: C

fskilnik · Answer

?\,\, = \,\,P\left( {{\rm{get}}\,\,{\rm{a}}\,\,{\rm{pair}}} \right)

\left. \matrix{
 {\rm{Total}}\,\,:\,\,\,C\left( {52,2} \right) = {{52 \cdot 51} \over 2} = 26 \cdot 51\,\,\,{\rm{equiprobable}}\,\,{\rm{choices}}{\kern 1pt} \,\, \hfill \cr 
 {\rm{Favorable}}\,\,:\,\,\,\underbrace {\,C\left( {4,2} \right)}_{2\,\,{\rm{suits}}} \cdot \underbrace {C\left( {13,1} \right)}_{{\rm{card}}\,{\rm{to}}\,{\rm{get}}\,{\rm{twice}}} = 6 \cdot 13\,\,{\rm{choices}} \hfill \cr} \right\}\,\,\,\,\, \Rightarrow \,\,\,\,\,?\,\, = \,\,{{6 \cdot 13} \over {26 \cdot 51}}\,\, = \,\,{1 \over {17}}

The correct answer is (C).

We follow the notations and rationale taught in the  GMATH  method.

Regards, 
Fabio.

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What is the probability that you get a pair when picking the top two c

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What is the probability that you get a pair when picking the top two c

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