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09 Mar 2020, 20:45
Kudos
Bookmarks
E
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:
13% (01:58) correct
87%
(03:02)
wrong
based on 54
sessions
History
Date
Time
Result
Not Attempted Yet
Joey and his friends played a new game that uses 20 playing cards made up of 2 suits of 10 cards each. The cards in each suit have values from 1 to 10. If Joey turns over 4 cards, what is the probability that he will find at least one pair of cards having the same value?
A) 3/5
B) 12/19
C) 16/19
D) 99/323
E) 224/323
A) 3/5
B) 12/19
C) 16/19
D) 99/323
E) 224/323
09 Mar 2020, 22:57
Kudos
Bookmarks
costcosized
Opposite probability through permutation
So, TWO sets of 10 each..
Way to choose two cards out of 20 = 20*19*18*17
Way to choose the cards such that all are different...
First can be any of 20 cards. The next can be any of the remaining 19 but the pair of the first chosen, so 19-1 or 18 ways. Similarly next in 16 and next in 14 ways.
Total ways = 20*18*16*14
Probability = \(\frac{20*18*16*14}{20*19*18*17}=\frac{16*14}{19*17}=\frac{224}{323}\)
So our probability = \(1-\frac{224}{323}=\frac{99}{323}\)
another way
\(C^4_{10}\) - ways to choose 4 out of 10 different values/cards
\(2^4\) - Now 4 cards can be chosen from 2 suits
\(C^4_{20}\) - total ways to choose 4 cards out of 20.
Opposite probability that is all 4 cards will be different \(\frac{C^4_{10}*2^4}{C^4_{20}}=\frac{224}{323}\).
as seen above \(1- \frac{224}{323}=\frac{99}{323}\)
Choosing the actual probability
For this you will have take each case and in turn wasting time in a lengthy process
OA will be D and not E
10 Mar 2020, 02:20
Kudos
Bookmarks
Great explanation.
