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Penny, Leonard, and Sheldon each have one try to make a basket from ha

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Bunuel
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Penny, Leonard, and Sheldon each have one try to make a basket from ha [#permalink] New post   05 Feb 2015, 09:20
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Penny, Leonard, and Sheldon each have one try to make a basket from half court. If their individual probabilities of making the basket are 1/4, 1/8, and 1/3, respectively, what is the probability that Penny and Leonard, but not Sheldon, will make the basket?

A. 1/96
B. 1/48
C. 1/24
D. 1/12
E. 17/24


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B
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sterling19
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Re: Penny, Leonard, and Sheldon each have one try to make a basket from ha [#permalink] New post   05 Feb 2015, 17:29
Sheldon may need more force: https://www.youtube.com/watch?v=4QPetCN6b6Q

P(Sheldon misses) = 1 - P(Sheldon makes) = 1 - 1/3 = 2/3
P(Penny makes AND Leonard makes AND Sheldon misses) = 1/4 * 1/8 * 2/3 = 2/96 = 1/48
The correct answer is B.
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Re: Penny, Leonard, and Sheldon each have one try to make a basket from ha [#permalink] New post   05 Feb 2015, 18:40
Bunuel wrote:
Penny, Leonard, and Sheldon each have one try to make a basket from half court. If their individual probabilities of making the basket are 1/4, 1/8, and 1/3, respectively, what is the probability that Penny and Leonard, but not Sheldon, will make the basket?

A. 1/96
B. 1/48
C. 1/24
D. 1/12
E. 17/24


Kudos for a correct solution.


(Probability Penny is successful) * (Probability Leonard is successful) * (Probability Sheldon is unsuccessful) =
(1/4)*(1/8)*(2/3) = 1/48

Answer: B

My question is how come we don't multiply (1/48) by 3? Order doesn't matter in this case, so don't we have to account for the number of ways?
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Re: Penny, Leonard, and Sheldon each have one try to make a basket from ha [#permalink] New post   05 Feb 2015, 18:47
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peachfuzz wrote:
Bunuel wrote:
Penny, Leonard, and Sheldon each have one try to make a basket from half court. If their individual probabilities of making the basket are 1/4, 1/8, and 1/3, respectively, what is the probability that Penny and Leonard, but not Sheldon, will make the basket?

A. 1/96
B. 1/48
C. 1/24
D. 1/12
E. 17/24


Kudos for a correct solution.


(Probability Penny is successful) * (Probability Leonard is successful) * (Probability Sheldon is unsuccessful) =
(1/4)*(1/8)*(2/3) = 1/48

Answer: B

My question is how come we don't multiply (1/48) by 3? Order doesn't matter in this case, so don't we have to account for the number of ways?


hi peachfuzz,
we multiply by 3 or required number in situations where u are making arrangements or combinations.

ans B..
1/4*1/8*2/3=1/48
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Bunuel
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Re: Penny, Leonard, and Sheldon each have one try to make a basket from ha [#permalink] New post   09 Feb 2015, 05:41
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Bunuel wrote:
Penny, Leonard, and Sheldon each have one try to make a basket from half court. If their individual probabilities of making the basket are 1/4, 1/8, and 1/3, respectively, what is the probability that Penny and Leonard, but not Sheldon, will make the basket?

A. 1/96
B. 1/48
C. 1/24
D. 1/12
E. 17/24


Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Correct Answer: B

Solution: B. Since we have independent probabilities, we simply have to multiply to figure out combined probability. Since we are looking for Sheldon to not make the basket, his probability is 1-(1/3), or 2/3. Our equation is (1/4) * (1/8) * (2/3) = (2/96 ) = (1/48). The answer is B.
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Re: Penny, Leonard, and Sheldon each have one try to make a basket from ha [#permalink] New post   16 Jan 2024, 18:26
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Re: Penny, Leonard, and Sheldon each have one try to make a basket from ha [#permalink]
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