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Four cards are chosen from a standard deck: two aces (one of Spades, a

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Four cards are chosen from a standard deck: two aces (one of Spades, a  [#permalink]

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New post 20 Feb 2019, 08:28
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  65% (hard)

Question Stats:

38% (02:06) correct 62% (03:00) wrong based on 21 sessions

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GMATH practice exercise (Quant Class 19)

Four cards are chosen from a standard deck: two aces (one of Spades, another of Hearts) and two kings (one of Spades, another of Hearts). The aces are considered as -1 (Spades) and 1 (Hearts), while the kings are considered as -2 (Spades) and 2 (Hearts). If two different cards among these four are randomly chosen, and their corresponding numerical values are multiplied together, which of the following is closest to the probability that the product obtained is negative or odd (or both)?

(A) 17%
(B) 50%
(C) 56%
(D) 67%
(E) 75%

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Re: Four cards are chosen from a standard deck: two aces (one of Spades, a  [#permalink]

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New post 20 Feb 2019, 09:18
fskilnik wrote:
GMATH practice exercise (Quant Class 19)

Four cards are chosen from a standard deck: two aces (one of Spades, another of Hearts) and two kings (one of Spades, another of Hearts). The aces are considered as -1 (Spades) and 1 (Hearts), while the kings are considered as -2 (Spades) and 2 (Hearts). If two different cards among these four are randomly chosen, and their corresponding numerical values are multiplied together, which of the following is closest to the probability that the product obtained is negative or odd (or both)?

(A) 17%
(B) 50%
(C) 56%
(D) 67%
(E) 75%


The following cases are possible:
(-1)(1) = -1
(-1)(-2) = 2
(-1)(2) = -2
(1)(-2) = -2
(1)(2) = 2
(-2)(2) = -4
Of the 6 cases above, the 4 in blue each yield a product that is negative, odd or both:
4/6 = 2/3 ≈ 67%

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Re: Four cards are chosen from a standard deck: two aces (one of Spades, a  [#permalink]

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New post 20 Feb 2019, 09:22
Let’s list out all the pairs where the product is odd or megative or both:
(-1,1),(-1,2),(-2,1),(-2,2),(1,-1),(1,-2),(2,-1),(2,-2) = 8 pairs

Total number of pairs = 12 pairs ( 4 cards are available when you pick for the first time and 3 are left when you pick for the second time so 4x3 = 12 pairs are pssible in total).

So probability = 8/12 = 67%

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Re: Four cards are chosen from a standard deck: two aces (one of Spades, a  [#permalink]

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New post 20 Feb 2019, 11:38
fskilnik wrote:
GMATH practice exercise (Quant Class 19)

Four cards are chosen from a standard deck: two aces (one of Spades, another of Hearts) and two kings (one of Spades, another of Hearts). The aces are considered as -1 (Spades) and 1 (Hearts), while the kings are considered as -2 (Spades) and 2 (Hearts). If two different cards among these four are randomly chosen, and their corresponding numerical values are multiplied together, which of the following is closest to the probability that the product obtained is negative or odd (or both)?

(A) 17%
(B) 50%
(C) 56%
(D) 67%
(E) 75%


since its P or OR so we need to add cases:
case 1: negative : (-1*1),(-2*1),(-1*2),(-2*2) ; 4 cases
case 2 : odd : ( -1*1),(-2*-1)*(-2*1)*(2*-1)*(2*1) : 5 cases
and odd & -ve : ( -1*1),(-2*1) , ( 2*-1) ; 3 cases

total ; 4+5+3 = 12 cases
and we have total multiple cases of ( -1,1) and ( 2,-2 ) ; which can make total of 8 pairs
so P would be 8/12 ; 67%
IMO D
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Re: Four cards are chosen from a standard deck: two aces (one of Spades, a  [#permalink]

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New post 20 Feb 2019, 11:58
fskilnik wrote:
GMATH practice exercise (Quant Class 19)

Four cards are chosen from a standard deck: two aces (one of Spades, another of Hearts) and two kings (one of Spades, another of Hearts). The aces are considered as -1 (Spades) and 1 (Hearts), while the kings are considered as -2 (Spades) and 2 (Hearts). If two different cards among these four are randomly chosen, and their corresponding numerical values are multiplied together, which of the following is closest to the probability that the product obtained is negative or odd (or both)?

(A) 17%
(B) 50%
(C) 56%
(D) 67%
(E) 75%

\(\left\{ { - 2, - 1,1,2} \right\}\,\, \to \,\,{\rm{two}}\,{\rm{different}}\,\,{\rm{chosen}}\)

\(? = P\left( {{\rm{odd}}\,\,{\rm{or}}\,\,{\rm{negative}}\,\,{\rm{product}}} \right)\)

\({\rm{6}}\,\,{\rm{equiprobable}}\,\,{\rm{outcomes}}\,\,:\,\,\,\left\{ \matrix{
\,\left\{ { - 2, - 1} \right\},\left\{ {1,2} \right\}\,\,\,\,\left( {{\rm{product}}\,\,2} \right) \hfill \cr
\,\left\{ { - 2,1} \right\},\left\{ { - 1,2} \right\}\,\,\,\,\left( {{\rm{product}}\,\, - 2} \right)\,\,\,::\,\,\,2\,\,{\rm{favorable}}\,\,{\rm{outcomes}} \hfill \cr
\,\left\{ { - 2,2} \right\}\,\,\,\,\left( {{\rm{product}}\,\, - 4} \right)\,\,\,::\,\,\,1\,\,{\rm{favorable}}\,\,{\rm{outcome}} \hfill \cr
\,\left\{ { - 1,1} \right\}\,\,\,\,\left( {{\rm{product}}\,\, - 1} \right)\,\,\,::\,\,\,1\,\,{\rm{favorable}}\,\,{\rm{outcome}} \hfill \cr} \right.\)

\(?\,\, = \,\,{{2 + 1 + 1} \over 6}\,\, = \,\,{2 \over 3}\,\, \cong \,\,67\%\)


The correct answer is (D).


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

Regards,
Fabio.
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Our high-level "quant" preparation starts here: https://gmath.net

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Re: Four cards are chosen from a standard deck: two aces (one of Spades, a   [#permalink] 20 Feb 2019, 11:58
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