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27 Aug 2017, 21:42
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
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Question Stats:
41% (02:33) correct
59%
(02:46)
wrong
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A lottery winner from State F must match, in any order, 6 balls randomly chosen from a single pool of balls numbered from 1 to 50. A lottery winner from State G must match, in any order, 5 balls randomly chosen from a first pool of balls numbered from 1 to 50 AND a “megaball,” randomly chosen from a second set of balls numbered from 1 to 50. The number of winning combinations in a single drawing of the lottery in State G is what percentage greater than the number of winning combinations in a single drawing of the lottery in State F?
A. 11%
B. 85%
C. 111%
D. 567%
E. 667%
A. 11%
B. 85%
C. 111%
D. 567%
E. 667%
27 Aug 2017, 22:04
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For F: Total combination is \(50C_6\) = \(\frac{50*49*48*47*46*45}{6*5*4*3*2*1}\)
For G: Total combination is \(50C_5\)*\(5C_1\) = \(\frac{50*49*48*47*46}{5*4*3*2*1}\)*\(\frac{50}{1}\)
% greater = \(\frac{change}{smaller}\)*100
change = \(\frac{50*49*48*47*46}{5*4*3*2*1}\)*\(\frac{50}{1}\) - \(\frac{50*49*48*47*46*45}{6*5*4*3*2*1}\)
= \(\frac{50*49*48*47*46}{5*4*3*2*1}\) (50 - \(\frac{45}{6}\))
% greater = \(\frac{\frac{50*49*48*47*46}{5*4*3*2*1}(50 - \frac{45}{6})}{\frac{50*49*48*47*46*45}{6*5*4*3*2*1}}\)*100
= \(\frac{50 - \frac{45}{6}}{\frac{45}{6}}\)*100
= \(\frac{50*6 - 45}{45}\)*100
=566.66%
Thus Option D
For G: Total combination is \(50C_5\)*\(5C_1\) = \(\frac{50*49*48*47*46}{5*4*3*2*1}\)*\(\frac{50}{1}\)
% greater = \(\frac{change}{smaller}\)*100
change = \(\frac{50*49*48*47*46}{5*4*3*2*1}\)*\(\frac{50}{1}\) - \(\frac{50*49*48*47*46*45}{6*5*4*3*2*1}\)
= \(\frac{50*49*48*47*46}{5*4*3*2*1}\) (50 - \(\frac{45}{6}\))
% greater = \(\frac{\frac{50*49*48*47*46}{5*4*3*2*1}(50 - \frac{45}{6})}{\frac{50*49*48*47*46*45}{6*5*4*3*2*1}}\)*100
= \(\frac{50 - \frac{45}{6}}{\frac{45}{6}}\)*100
= \(\frac{50*6 - 45}{45}\)*100
=566.66%
Thus Option D
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28 Aug 2017, 02:47
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HolaMaven
For F: Total combination is \(50C_6\) = \(\frac{50*49*48*47*46*45}{6*5*4*3*2*1}\)
For G: Total combination is \(50C_5\)*\(5C_1\) = \(\frac{50*49*48*47*46}{5*4*3*2*1}\)*\(\frac{50}{1}\)
% greater = \(\frac{change}{smaller}\)*100
change = \(\frac{50*49*48*47*46}{5*4*3*2*1}\)*\(\frac{50}{1}\) - \(\frac{50*49*48*47*46*45}{6*5*4*3*2*1}\)
= \(\frac{50*49*48*47*46}{5*4*3*2*1}\) (50 - \(\frac{45}{6}\))
% greater = \(\frac{\frac{50*49*48*47*46}{5*4*3*2*1}(50 - \frac{45}{6})}{\frac{50*49*48*47*46*45}{6*5*4*3*2*1}}\)*100
= \(\frac{50 - \frac{45}{6}}{\frac{45}{6}}\)*100
= \(\frac{50*6 - 45}{45}\)*100
=566.66%
Thus Option D
For G: Total combination is \(50C_5\)*\(5C_1\) = \(\frac{50*49*48*47*46}{5*4*3*2*1}\)*\(\frac{50}{1}\)
% greater = \(\frac{change}{smaller}\)*100
change = \(\frac{50*49*48*47*46}{5*4*3*2*1}\)*\(\frac{50}{1}\) - \(\frac{50*49*48*47*46*45}{6*5*4*3*2*1}\)
= \(\frac{50*49*48*47*46}{5*4*3*2*1}\) (50 - \(\frac{45}{6}\))
% greater = \(\frac{\frac{50*49*48*47*46}{5*4*3*2*1}(50 - \frac{45}{6})}{\frac{50*49*48*47*46*45}{6*5*4*3*2*1}}\)*100
= \(\frac{50 - \frac{45}{6}}{\frac{45}{6}}\)*100
= \(\frac{50*6 - 45}{45}\)*100
=566.66%
Thus Option D
HolaMaven , can you explain the one that I highlighted?
If 5C1 then it should be \(\frac{5}{1}\).
