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. 12/2,652
B. 16/2,652
C. 1/17
D. 1/13
E. 1/2
The OA is C
Source: Veritas Prep
What is the probability that you get a pair when picking the
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- Brent@GMATPrepNow
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P(select pair) = P(1st card is ANY card AND 2nd card matches 1st card)swerve wrote: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. 12/2,652
B. 16/2,652
C. 1/17
D. 1/13
E. 1/2
The OA is C
Source: Veritas Prep
= 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.
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Brent
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$$?\,\, = \,\,P\left( {{\rm{get}}\,\,{\rm{a}}\,\,{\rm{pair}}} \right)$$swerve wrote: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. 12/2,652
B. 16/2,652
C. 1/17
D. 1/13
E. 1/2
Source: Veritas Prep
$$\left. \matrix{
{\rm{Total}}\,\,:\,\,\,C\left( {52,2} \right) = {{52 \cdot 51} \over 2} = 26 \cdot 51\,\,\,{\rm{equiprobable}}\,\,{\rm{choices}}{\kern 1pt} \,\, \hfill \cr
{\rm{Favorable}}\,\,:\,\,\,\underbrace {\,C\left( {4,2} \right)}_{2\,\,{\rm{suits}}} \cdot \underbrace {C\left( {13,1} \right)}_{{\rm{card}}\,{\rm{to}}\,{\rm{get}}\,{\rm{twice}}} = 6 \cdot 13\,\,{\rm{choices}} \hfill \cr} \right\}\,\,\,\,\, \Rightarrow \,\,\,\,\,?\,\, = \,\,{{6 \cdot 13} \over {26 \cdot 51}}\,\, = \,\,{1 \over {17}}$$
The correct answer is therefore (C).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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