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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
41% (02:54) correct
59%
(02:33)
wrong
based on 114
sessions
History
Date
Time
Result
Not Attempted Yet
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
(A) 1/6
(B) 1/8
(C) 1/9
(D) 1/10
(E) 1/12
My answer differs from the OA.
24 Dec 2012, 03:35
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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
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
