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Three letters are randomly selected one at a time without replacement
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22 Jun 2021, 07:52
22 Jun 2021, 07:52
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Three letters are randomly selected one at a time without replacement from {A, B, C, D, E, F, G, H}. What is the probability that the 3 letters are removed in consecutive alphabetical order (e.g., select B, then C then D, or select F, then G then H or ....etc)?
A) 1/336
B) 1/112
C) 1/56
D) 1/8
E) 1/7
A) 1/336
B) 1/112
C) 1/56
D) 1/8
E) 1/7
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Re: Three letters are randomly selected one at a time without replacement
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22 Jun 2021, 10:36
22 Jun 2021, 10:36
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GreenlightTestPrep wrote:
Three letters are randomly selected one at a time without replacement from {A, B, C, D, E, F, G, H}. What is the probability that the 3 letters are removed in consecutive alphabetical order (e.g., select B, then C then D, or select F, then G then H or ....etc)?
A) 1/336
B) 1/112
C) 1/56
D) 1/8
E) 1/7
A) 1/336
B) 1/112
C) 1/56
D) 1/8
E) 1/7
Possible combinations = (ABC), (BCD), (CDE), (DEF), (EFG), (FGH)
Total outcomes = (8)(7)(6)
Required Prob. = \(\frac{1}{56}\)
Hence, option C
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: Three letters are randomly selected one at a time without replacement
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23 Jun 2021, 05:58
23 Jun 2021, 05:58
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GreenlightTestPrep wrote:
Three letters are randomly selected one at a time without replacement from {A, B, C, D, E, F, G, H}. What is the probability that the 3 letters are removed in consecutive alphabetical order (e.g., select B, then C then D, or select F, then G then H or ....etc)?
A) 1/336
B) 1/112
C) 1/56
D) 1/8
E) 1/7
A) 1/336
B) 1/112
C) 1/56
D) 1/8
E) 1/7
KarunMendiratta has provided a nice solution that uses counting techniques.
Here's a solution that involves probability rules
First recognize that the 1st selected letter must be A, B, C, D, E or F, otherwise it will be impossible to have three consecutive letters (e.g., if we select G first, then the next two letters would need to be H and I, but the letters only go up to H).
So,....
P(the 3 selected letters are in order) = P(1st selected letter is A, B, C, D, E or F AND the 2nd letter is the letter than follows the 1st selected letter AND the 3rd letter is the letter than follows the 2nd selected letter)
= P(1st selected letter is A, B, C, D, E or F) x P(2nd letter is the letter than follows the 1st selected letter) x P(3rd letter is the letter than follows the 2nd selected letter)
= 6/8 x 1/7 x 1/6
= (6)(1)(1)/(8)(7)(6)
= (1)(1)/(8)(7)
= 1/56
Answer: C
