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A three-letter code is formed using the letters A-L, such th
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09 Oct 2017, 23:10
09 Oct 2017, 23:10
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Question Stats:
43% (02:43) correct
56% (02:27) wrong based on 32 sessions
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A three-letter code is formed using the letters A-L, such that no letter is used more than once. What is the probability that the code will have a string of three consecutive letters (e.g. A-B-C, F-E-D)?
(A) 1/55
(B) 1/66
(C) 2/17
(D) 1/110
(E) 2/55
Kudos for correct solution.
(A) 1/55
(B) 1/66
(C) 2/17
(D) 1/110
(E) 2/55
Kudos for correct solution.
Re: A three-letter code is formed using the letters A-L, such th
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10 Oct 2017, 04:52
10 Oct 2017, 04:52
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Bunuel wrote:
A three-letter code is formed using the letters A-L, such that no letter is used more than once. What is the probability that the code will have a string of three consecutive letters (e.g. A-B-C, F-E-D)?
(A) 1/55
(B) 1/66
(C) 2/17
(D) 1/110
(E) 2/55
Kudos for correct solution.
(A) 1/55
(B) 1/66
(C) 2/17
(D) 1/110
(E) 2/55
Kudos for correct solution.
Now let us arrange three letter code in the form (A-B-C) using the letters from A to L
i.e we have = ABC, BCD, CDE, DEF, EFG, FGH, GHI, HIJ, IJK, JKL
i.e we have 10 options.
Similarly we can write the three letter code in the form (F-E-D) i.e in opposite direction and we still have 10 options.
Therefore we have a total of 20 options, which is the favourable outcomes
Now Total outcomes = 12 * 11 *10 (Letters from A to L is 12 and here order matters, so we cannot use combination )
therefore\(Probability =\frac{No. of favorable outcomes}{Total no. of outcomes}\)
or Probability = \(\frac{20}{(12*11*10}) = \frac{20}{1320} = \frac{1}{66}\)
Re: A three-letter code is formed using the letters A-L, such th
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19 May 2022, 11:18
19 May 2022, 11:18
But the question said, "such that no letter is used more than once".
