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There are y bricks in a row. If one brick is to be selected
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12 Nov 2020, 10:44
12 Nov 2020, 10:44
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Question Stats:
84% (00:35) correct
15% (02:34) wrong based on 13 sessions
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There are y bricks in a row. If one brick is to be selected at random, the probability that it will be cracked is \(\frac{3}{11}\). In terms of y, how many of the bricks are not cracked?
A. y/11
B. 8y/11
C. 11y/8
D. 3y/11
E. 11y
Source: 320 GRE Math Questions
A. y/11
B. 8y/11
C. 11y/8
D. 3y/11
E. 11y
Source: 320 GRE Math Questions
Re: There are y bricks in a row. If one brick is to be selected
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18 Nov 2020, 08:34
18 Nov 2020, 08:34
1
Given:
There are y bricks therefore possibility of selecting a cracked brick is \(\frac{3}{11}\) or \(\frac{3}{11}\) y
hence probability of selecting uncracked brick is \(y-\frac{3}{11}\) y that reduces to \(\frac{11}{11}\) \(y - \frac{3}{11}\) y
= \(\frac{8}{11}\) y
There are y bricks therefore possibility of selecting a cracked brick is \(\frac{3}{11}\) or \(\frac{3}{11}\) y
hence probability of selecting uncracked brick is \(y-\frac{3}{11}\) y that reduces to \(\frac{11}{11}\) \(y - \frac{3}{11}\) y
= \(\frac{8}{11}\) y
gmatclubot
Re: There are y bricks in a row. If one brick is to be selected [#permalink]
18 Nov 2020, 08:34
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