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Probability - GMAT Practice Questions

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Probability - GMAT Practice Questions [#permalink]

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New post 27 Aug 2010, 18:39
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Question Stats:

56% (01:31) correct 44% (00:40) wrong based on 18 sessions

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Question: James and Colleen are playing basketball. The probability of James missing a shot is x, and the probability of Colleen not making a shot is y. If they each take 2 shots. What is the probability that they both make at least 1 shot apiece?

A) \(1 - (x^2*y^2)\)
B) \((1-x^2) (1-y^2)\)
C) \((1-(1-x)^2) (1- (1-y)^2)\)
D) \((1-(1-x)^2) ((1-y)^2)\)
E) \((1-(1-y^2)) (1-x^2)\)

I could not solve this problem and I wanted to arrive at the answer using the algebraic route and by not picking numbers.
[Reveal] Spoiler: OA

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Re: Probability - GMAT Practice Questions [#permalink]

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New post 27 Aug 2010, 19:35
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ezhilkumarank wrote:
Question: James and Colleen are playing basketball. The probability of James missing a shot is x, and the probability of Colleen not making a shot is y. If they each take 2 shots. What is the probability that they both make at least 1 shot apiece?

A) \(1 - (x^2*y^2)\)
B) \((1-x^2) (1-y^2)\)
C) \((1-(1-x)^2) (1- (1-y)^2)\)
D) \((1-(1-x)^2) ((1-y)^2)\)
E) \((1-(1-y^2)) (1-x^2)\)

I could not solve this problem and I wanted to arrive at the answer using the algebraic route and by not picking numbers.


The probability James misses both shots is x*x = x^2. So the probability James makes at least one shot is 1 - x^2. Similarly, the probability Colleen makes at least one shot is 1 - y^2. To find the probability they both make at least one shot, we multiply the probability that each makes at least one shot: (1 - x^2)(1 - y^2).
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Senior Manager
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Status: Time to step up the tempo
Joined: 24 Jun 2010
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Kudos [?]: 195 [0], given: 50

Re: Probability - GMAT Practice Questions [#permalink]

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New post 27 Aug 2010, 20:16
IanStewart wrote:
ezhilkumarank wrote:
Question: James and Colleen are playing basketball. The probability of James missing a shot is x, and the probability of Colleen not making a shot is y. If they each take 2 shots. What is the probability that they both make at least 1 shot apiece?

A) \(1 - (x^2*y^2)\)
B) \((1-x^2) (1-y^2)\)
C) \((1-(1-x)^2) (1- (1-y)^2)\)
D) \((1-(1-x)^2) ((1-y)^2)\)
E) \((1-(1-y^2)) (1-x^2)\)

I could not solve this problem and I wanted to arrive at the answer using the algebraic route and by not picking numbers.


The probability James misses both shots is x*x = x^2. So the probability James makes at least one shot is 1 - x^2. Similarly, the probability Colleen makes at least one shot is 1 - y^2. To find the probability they both make at least one shot, we multiply the probability that each makes at least one shot: (1 - x^2)(1 - y^2).



Thanks IanStewart. +1 from me.
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Re: Probability - GMAT Practice Questions   [#permalink] 27 Aug 2010, 20:16
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