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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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Updated on: 07 Aug 2019, 07:27
Question Stats:
44% (02:32) correct 56% (02:49) wrong based on 52 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.
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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.




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Re: Amy has two decks of 52 cards each: Deck 1 and Deck 2. She takes 8 bla
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24 Dec 2012, 02:35
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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Re: Amy has two decks of 52 cards each: Deck 1 and Deck 2. She takes 8 bla
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24 Dec 2012, 03:13
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 Your solution is correct answer should be 1/10.
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Re: Amy has two decks of 52 cards each: Deck 1 and Deck 2. She takes 8 bla
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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 (5,52) 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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Re: Amy has two decks of 52 cards each: Deck 1 and Deck 2. She takes 8 bla
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07 Aug 2019, 06:47
Hi Bunuel, kindly edit the OA here. Posted from my mobile device



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Re: Amy has two decks of 52 cards each: Deck 1 and Deck 2. She takes 8 bla
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07 Aug 2019, 07:28
omkartadsare wrote: 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
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07 Aug 2019, 07:28




