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21 Jun 2022, 15:15
Kudos
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
67% (02:31) correct
33%
(02:34)
wrong
based on 204
sessions
History
Date
Time
Result
Not Attempted Yet
A word consists of 10 letters; 6 consonants and 4vowels. Three letters are chosen at random. What is the probability that more than one vowel will be selected ?
A. 1/7
B. 1/3
C. 17/42
D. 4/5
E. 6/7
A. 1/7
B. 1/3
C. 17/42
D. 4/5
E. 6/7
GMATCoachBen
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Expert reply
21 Jun 2022, 17:05
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Bunuel
For probability, we need to find: \(\frac{ # of Desired Combinations }{ # of Total Combinations}\)
We use the combinations formula, because order doesn't matter here: nCr = \(\frac{n!}{r!(n−r)!}\). (nCr means "n choose r" — the # of combinations when out of "n" items we choose "r" items)
For the Desired Combinations, there are two ways to get "more than one vowel"; we can have either:
A) 2 vowels AND 1 consonant OR
B) 3 vowels AND 0 consonants
Note that "OR" means we "add", and "AND" means we multiply.
Step 1:
Combinations with 2 vowels AND 1 consonant (we are choosing 2 vowels out of 4, and 1 consonant out of 6) = 4C2 * 6C1 = \(\frac{4*3}{2}\) * 6 = 36
Step 2:
Combinations with 3 vowels (AND 0 consonants) = 4C3 = 4 (Note: there is always only 1 combination when we choose 0 items, so we don't need to do a calculation for the 0 consonants)
Step 3:
Total Combinations, choosing 3 letters out of 10 letters = 10C3 = \(\frac{10*9*8}{(3*2)}\) = 120
Step 4:
Probability = \(\frac{Combinations With 2 or 3 vowels }{ Total Combinations}\) = \(\frac{36 + 4 }{ 120}\) = 1/3
General Discussion
21 Jun 2022, 20:25
Kudos
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Instead to attack directly what he asks, I calculated what he doesn’t ask and subtracted from the sample set that is always 1. It’s faster.
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