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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.
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.
27 Aug 2010, 20:35
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ezhilkumarank
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.
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).
General Discussion
27 Aug 2010, 21:16
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IanStewart
ezhilkumarank
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.
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.
