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24 Nov 2021, 08:17
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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. \(\frac{1}{60}\)
B. \(\frac{1}{30}\)
C. \(\frac{1}{20}\)
D. \(\frac{1}{6}\)
E. \(\frac{1}{3}\)
A. \(\frac{1}{60}\)
B. \(\frac{1}{30}\)
C. \(\frac{1}{20}\)
D. \(\frac{1}{6}\)
E. \(\frac{1}{3}\)
24 Nov 2021, 08:17
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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
29 Sep 2024, 13:41
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Thanks for your explanation.
I want to know, why in this case order matter and we need to calculate total ways to choose 3 cards using Permutation instead of a regular combinatory
I want to know, why in this case order matter and we need to calculate total ways to choose 3 cards using Permutation instead of a regular combinatory
