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Bunuel
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Every year Taylor goes to the same carnival, and he attempts to shoot 06 Apr 2016, 07:04
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66% (01:26) correct 34% (01:29) wrong based on 216 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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Every year Taylor goes to the same carnival, and he attempts to shoot 06 Apr 2016, 08:48
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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

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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Every year Taylor goes to the same carnival, and he attempts to shoot 18 Jul 2017, 06:32
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FightToSurvive
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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Why do we need to multiply 3 to (1*9*9/1000)?

Thanks
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Every year Taylor goes to the same carnival, and he attempts to shoot 05 May 2019, 10:54
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Probability of tony making the basket - 1/10
Probability of tony not making the basket - 9/10

Exactly one basket in 3 tries = \((\frac{1}{10}*\frac{9}{10}*\frac{9}{10})*3\) = \(\frac{3^5}{10^3}\)

A is the answer.
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Every year Taylor goes to the same carnival, and he attempts to shoot 12 May 2019, 06:22
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Gymat
FightToSurvive
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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Why do we need to multiply 3 to (1*9*9/1000)?

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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
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Every year Taylor goes to the same carnival, and he attempts to shoot Updated on: 30 Jan 2026, 13:13
Originally posted by luisdicampo on 30 Jan 2026, 09:51.
Last edited by Bunuel on 30 Jan 2026, 13:13, edited 1 time in total.
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Deconstructing the Question
\(P(\text{miss})=9/10\), so \(P(\text{make})=1/10\).
Taylor takes 3 independent shots.
We want \(P(\text{exactly 1 make})\).

Step-by-step
Pick one specific pattern (make, miss, miss):
\((1/10)(9/10)(9/10)\)

The make can occur in any of the 3 shot positions:
\(C(3,1)=3\)

So,
\(P=3\cdot(1/10)(9/10)(9/10)=\frac{3\cdot 81}{10^3}=\frac{243}{10^3}\)

Since \(243=3^5\), we get:
\(P=\frac{3^5}{10^3}\)

Answer: (A) 3^5 / 10^3
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