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# Amy has two decks of 52 cards each: Deck 1 and Deck 2. She takes 8 bla

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Joined: 14 Jun 2011
Posts: 15
Amy has two decks of 52 cards each: Deck 1 and Deck 2. She takes 8 bla  [#permalink]

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Updated on: 07 Aug 2019, 07:27
3
6
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Difficulty:

95% (hard)

Question Stats:

45% (02:32) correct 55% (02:51) wrong based on 51 sessions

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Amy has two decks of 52 cards each: Deck 1 and Deck 2. She takes 8 black cards from Deck 2 and adds them to Deck 1 and shuffles it thoroughly. She now picks a card from the newly formed pack of cards. If the probability of either picking a red ace or a king from the newly formed pack is greater than 1/8, what is the probability that Amy picks a black king or a red Jack from the new pack?

(A) 1/6
(B) 1/8
(C) 1/9
(D) 1/10
(E) 1/12

My answer differs from the OA.

Originally posted by tabsang on 24 Dec 2012, 02:33.
Last edited by Bunuel on 07 Aug 2019, 07:27, edited 3 times in total.
Edited the OA.
Intern
Joined: 14 Jun 2011
Posts: 15
Re: Amy has two decks of 52 cards each: Deck 1 and Deck 2. She takes 8 bla  [#permalink]

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24 Dec 2012, 02:35
5
1
Here's my approach:

Let "k" be the number of kings (black) that may be part of the 8 BLACK cards added to Deck 1.
No. of red aces in the new deck remains unchanged i.e. 2
No. of cards in the new deck = 52+8 = 60
No. of kings in the new deck = 4+k (2 Red kings + 2 Black kings + k Black kings)

Thus, probability of picking a red ace or a king is :

2/(52+8) + (4+k)/(52+8) = (6+k)/60

Now, since it's given that the above probability is greater than 1/8
We get,
(6+k)/60 > 1/8

i.e. k > 3/2
i.e. k > 1

k can only take 3 values (0 black kings or 1 black king or 2 black kings)
Thus, since k>1, k=2.

Now, the required probability of Amy picking a black king or a red jack from the new pack can be calculated as follows:

Total no. of black kings in the new pack = 4 (2 from Deck 1 + 2 from the 8 cards added to Deck 1)
Total no. of red Jacks in the new pack = 2

Thus,

4/60 + 2/60 = 6/60 = 1/10.

I got (D) as the answer.
Did anyone else get the same answer???

Cheers,
Taz
##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 64248
Re: Amy has two decks of 52 cards each: Deck 1 and Deck 2. She takes 8 bla  [#permalink]

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24 Dec 2012, 03:13
2
2
tabsang wrote:
Amy has two decks of 52 cards each: Deck 1 and Deck 2. She takes 8 black cards from Deck 2 and adds them to Deck 1 and shuffles it thoroughly. She now picks a card from the newly formed pack of cards. If the probability of either picking a red ace or a king from the newly formed pack is greater than 1/8, what is the probability that Amy picks a black king or a red Jack from the new pack?

(A) 1/6
(B) 1/8
(C) 1/9
(D) 1/10
(E) 1/12

Here's my approach:

Let "k" be the number of kings (black) that may be part of the 8 BLACK cards added to Deck 1.
No. of red aces in the new deck remains unchanged i.e. 2
No. of cards in the new deck = 52+8 = 60
No. of kings in the new deck = 4+k (2 Red kings + 2 Black kings + k Black kings)

Thus, probability of picking a red ace or a king is :

2/(52+8) + (4+k)/(52+8) = (6+k)/60

Now, since it's given that the above probability is greater than 1/8
We get,
(6+k)/60 > 1/8

i.e. k > 3/2
i.e. k > 1

k can only take 3 values (0 black kings or 1 black king or 2 black kings)
Thus, since k>1, k=2.

Now, the required probability of Amy picking a black king or a red jack from the new pack can be calculated as follows:

Total no. of black kings in the new pack = 4 (2 from Deck 1 + 2 from the 8 cards added to Deck 1)
Total no. of red Jacks in the new pack = 2

Thus,

4/60 + 2/60 = 6/60 = 1/10.

I got (D) as the answer.
Did anyone else get the same answer???

Cheers,
Taz

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Joined: 28 Feb 2012
Posts: 103
GPA: 3.9
WE: Marketing (Other)
Re: Amy has two decks of 52 cards each: Deck 1 and Deck 2. She takes 8 bla  [#permalink]

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26 Dec 2012, 01:26
Interesting questions, i had almost the same solution, but without using unknowns.

we know that there are 2 red aces, then probability of having one red ace and one black king is bigger than 1/8; 2/60+1/x > 1/8; 1/x>11/120, which is about 5,5/60.
Since we know taht there are 2 red Jacks, then there should be more than 3,5 (5,5-2) black kings. That is possible only when there are 4 black kings, so overall probability of having 2 red jacks and 4 black kings is 6/60=1/10

Hope that helps!
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Joined: 11 Jun 2019
Posts: 34
Location: India
Re: Amy has two decks of 52 cards each: Deck 1 and Deck 2. She takes 8 bla  [#permalink]

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07 Aug 2019, 06:47
Hi Bunuel, kindly edit the OA here.

Posted from my mobile device
Math Expert
Joined: 02 Sep 2009
Posts: 64248
Re: Amy has two decks of 52 cards each: Deck 1 and Deck 2. She takes 8 bla  [#permalink]

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07 Aug 2019, 07:28
Hi Bunuel, kindly edit the OA here.

Posted from my mobile device

____________________________
Done.
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Re: Amy has two decks of 52 cards each: Deck 1 and Deck 2. She takes 8 bla   [#permalink] 07 Aug 2019, 07:28