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James and Colleen are playing basketball. The probability of James mis

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James and Colleen are playing basketball. The probability of James mis [#permalink]

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New post 27 Aug 2010, 19:39
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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: James and Colleen are playing basketball. The probability of James mis [#permalink]

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New post 27 Aug 2010, 20:35
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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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Kudos [?]: 1953 [1], given: 6

Senior Manager
Senior Manager
User avatar
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 404

Kudos [?]: 249 [0], given: 50

Location: Milky way
Schools: ISB, Tepper - CMU, Chicago Booth, LSB
Re: James and Colleen are playing basketball. The probability of James mis [#permalink]

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New post 27 Aug 2010, 21:16
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: James and Colleen are playing basketball. The probability of James mis [#permalink]

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Kudos [?]: 273 [0], given: 0

Re: James and Colleen are playing basketball. The probability of James mis   [#permalink] 24 Sep 2017, 00:36
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