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

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ezhilkumarank
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James and Colleen are playing basketball. The probability of James mis [#permalink] 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.
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B
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IanStewart
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Re: James and Colleen are playing basketball. The probability of James mis [#permalink] 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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Re: James and Colleen are playing basketball. The probability of James mis [#permalink] 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] New post   20 Oct 2019, 05:13
Could someone tell me how the solution would change if the question asked for the probability that both make exactly one shot?
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Re: James and Colleen are playing basketball. The probability of James mis [#permalink] New post   30 Nov 2020, 11:34
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ezhilkumarank wrote:
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) \((1-(1-y^2)) (1-x^2)\)


Solution:

The probability that James will miss both of his shots is x * x = x^2. Thus, the probability that he will make at least 1 of his shots is 1 - x^2. Similarly, the probability that Colleen will miss both of her shots is y * y = y^2. Thus, the probability that she will make at least 1 of his shots is 1 - y^2. Therefore, the probability that they both make at least 1 shot apiece is (1 - x^2)(1 - y^2).

Answer: B
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Re: James and Colleen are playing basketball. The probability of James mis [#permalink] New post   13 Feb 2021, 19:29
icanhazmba wrote:
Could someone tell me how the solution would change if the question asked for the probability that both make exactly one shot?


Each person can miss either Shot#1 or Shot#2 but not both.

Probability of James missing exactly 1 shot = 2C1 * x(1-x)
Probability of Colleen missing exactly 1 shot = 2C1 * y(1-y)

Since these both are independent events, probability of both making exactly one shot = 4xy(1-x)(1-y)
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Re: James and Colleen are playing basketball. The probability of James mis [#permalink] New post   20 May 2022, 22:11
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Re: James and Colleen are playing basketball. The probability of James mis [#permalink]
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