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Every year Taylor goes to the same carnival, and he attempts to shoot

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Every year Taylor goes to the same carnival, and he attempts to shoot  [#permalink]

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New post 06 Apr 2016, 07:04
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Question Stats:

68% (01:40) correct 32% (01:40) wrong based on 53 sessions

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Every year Taylor goes to the same carnival, and he attempts to shoot a basketball into a small hoop, hoping that he will win a ten-foot tall teddy bear. If the probability that Taylor does not make a basket is 9/10, what is the probability that Taylor makes exactly one basket in three tries?

A. 3^5/10^3
B. 3^4/10^3
C. 3^3/10^3
D. 3^2/10^3
E. 3^4/10^4

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Re: Every year Taylor goes to the same carnival, and he attempts to shoot  [#permalink]

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New post 06 Apr 2016, 08:48
Every year Taylor goes to the same carnival, and he attempts to shoot a basketball into a small hoop, hoping that he will win a ten-foot tall teddy bear. If the probability that Taylor does not make a basket is 9/10, what is the probability that Taylor makes exactly one basket in three tries?

A. 3^5/10^3
B. 3^4/10^3
C. 3^3/10^3
D. 3^2/10^3
E. 3^4/10^4

Probability of basket = 1/10
Probability of no basket = 9/10

Required probability= 3* (1*9*9/1000)
= 3^5/10^3
A is the answer IMO.

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Re: Every year Taylor goes to the same carnival, and he attempts to shoot  [#permalink]

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New post 18 Jul 2017, 06:32
FightToSurvive wrote:
Every year Taylor goes to the same carnival, and he attempts to shoot a basketball into a small hoop, hoping that he will win a ten-foot tall teddy bear. If the probability that Taylor does not make a basket is 9/10, what is the probability that Taylor makes exactly one basket in three tries?

A. 3^5/10^3
B. 3^4/10^3
C. 3^3/10^3
D. 3^2/10^3
E. 3^4/10^4

Probability of basket = 1/10
Probability of no basket = 9/10

Required probability= 3* (1*9*9/1000)
= 3^5/10^3
A is the answer IMO.

Posted from my mobile device



Why do we need to multiply 3 to (1*9*9/1000)?

Thanks
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Re: Every year Taylor goes to the same carnival, and he attempts to shoot  [#permalink]

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New post 12 May 2019, 06:22
Gymat wrote:
FightToSurvive wrote:
Every year Taylor goes to the same carnival, and he attempts to shoot a basketball into a small hoop, hoping that he will win a ten-foot tall teddy bear. If the probability that Taylor does not make a basket is 9/10, what is the probability that Taylor makes exactly one basket in three tries?

A. 3^5/10^3
B. 3^4/10^3
C. 3^3/10^3
D. 3^2/10^3
E. 3^4/10^4

Probability of basket = 1/10
Probability of no basket = 9/10

Required probability= 3* (1*9*9/1000)
= 3^5/10^3
A is the answer IMO.

Posted from my mobile device



Why do we need to multiply 3 to (1*9*9/1000)?

Thanks


I think because you have three ways of realizing one basket and failing the other two: BASKET,NOT_BASKET,NOT_BASKET OR NOT_BASKET,BASKET,NOT_BASKET OR NOT_BASKET,NOT_BASKET,BASKET
hope that helps
GMAT Club Bot
Re: Every year Taylor goes to the same carnival, and he attempts to shoot   [#permalink] 12 May 2019, 06:22
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Every year Taylor goes to the same carnival, and he attempts to shoot

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