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22 Apr 2020, 11:37
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Difficulty:
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69% (01:51) correct
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A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6. If three cards are randomly selected one by one from the deck without replacement, what is the probability that the numbers on the cards are consecutive integers in increasing order?
A. 1/60
B. 1/30
C. 1/20
D. 1/6
E. 1/3
M37-45
A. 1/60
B. 1/30
C. 1/20
D. 1/6
E. 1/3
M37-45
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24 Nov 2021, 08:18
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Official Solution:
A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6. If three cards are randomly selected one by one from the deck without replacement, what is the probability that the numbers on the cards are consecutive integers in increasing order?
A. \(\frac{1}{60}\)
B. \(\frac{1}{30}\)
C. \(\frac{1}{20}\)
D. \(\frac{1}{6}\)
E. \(\frac{1}{3}\)
The number of ways to choose 3 numbers out of 6 when order matters is \(P^3_6=\frac{6!}{3!}=6*5*4=120\);
The number of ways to get 3 consecutive numbers in increasing order is 4: {1, 2, 3}, {2, 3, 4}, {3, 4, 5} and {4, 5, 6};
The probability is therefore, \(\frac{4}{120}=\frac{1}{30}\).
Answer: B
A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6. If three cards are randomly selected one by one from the deck without replacement, what is the probability that the numbers on the cards are consecutive integers in increasing order?
A. \(\frac{1}{60}\)
B. \(\frac{1}{30}\)
C. \(\frac{1}{20}\)
D. \(\frac{1}{6}\)
E. \(\frac{1}{3}\)
The number of ways to choose 3 numbers out of 6 when order matters is \(P^3_6=\frac{6!}{3!}=6*5*4=120\);
The number of ways to get 3 consecutive numbers in increasing order is 4: {1, 2, 3}, {2, 3, 4}, {3, 4, 5} and {4, 5, 6};
The probability is therefore, \(\frac{4}{120}=\frac{1}{30}\).
Answer: B
General Discussion
22 Apr 2020, 13:16
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Total number of possible arrangements of 3 cards = 6*5*4 = 120
Number of ways to select consecutive integers cards in increasing order = 4 [{1,2,3}, {2,3,4}, {3,4,5}, {4,5,6}]
Probability = 4/120 = 1/30
Number of ways to select consecutive integers cards in increasing order = 4 [{1,2,3}, {2,3,4}, {3,4,5}, {4,5,6}]
Probability = 4/120 = 1/30
Bunuel
