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James and Logan are taking batting practice. If their

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by AAPL » Mon Aug 12, 2019 4:17 am

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B

C

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E

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Difficulty—

Princeton Review

James and Logan are taking batting practice. If their individual probabilities of hitting a homerun are \(x\) and \(y\), respectively, then what is the probability that James will not hit a homerun but Logan will?

A. \(x - xy\)
B. \(x - y^2\)
C. \(y - xy\)
D. \(\frac{y}{x}\)
E. \(\frac{1}{x} - y\)

OA C
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by Brent@GMATPrepNow » Mon Aug 12, 2019 5:56 am
AAPL wrote:Princeton Review

James and Logan are taking batting practice. If their individual probabilities of hitting a homerun are \(x\) and \(y\), respectively, then what is the probability that James will not hit a homerun but Logan will?

A. \(x - xy\)
B. \(x - y^2\)
C. \(y - xy\)
D. \(\frac{y}{x}\)
E. \(\frac{1}{x} - y\)

OA C
If P(James hits home run) = x, then P(James does NOT hit home run) = 1-x
Given: P(Logan hits home run) = y

So, P(James will not hit a home run AND Logan hits a home run) = P(James will not hit a home run) X P(Logan hits a home run)
= (1-x) X (y)
= y - yx

Answer: C

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Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Scott@TargetTestPrep » Wed Aug 14, 2019 6:56 pm
AAPL wrote:Princeton Review

James and Logan are taking batting practice. If their individual probabilities of hitting a homerun are \(x\) and \(y\), respectively, then what is the probability that James will not hit a homerun but Logan will?

A. \(x - xy\)
B. \(x - y^2\)
C. \(y - xy\)
D. \(\frac{y}{x}\)
E. \(\frac{1}{x} - y\)

OA C
The probability that James will not hit a homerun but Logan will is (1 - x)(y) = y - yx.

Answer: C

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