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655-705 Level|   Probability|      
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There are 15 children in a schoolyard from which a team of 3 will be 09 Jun 2020, 11:13
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E
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Difficulty:

65% (hard)

Question Stats:

63% (02:40) correct 37% (02:47) wrong based on 220 sessions

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There are 15 children in a schoolyard from which a team of 3 will be chosen. One child is named Joey; another is named Anton. What is the probability that Joey is included in the team and Anton is excluded from the team?

A. 2/15
B. 39/182
C. 1/5
D. 1/4
E. 78/455
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E
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There are 15 children in a schoolyard from which a team of 3 will be 10 Jun 2020, 00:23
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As we want Joey to be included and Anton to be excluded from the team, we have to select 2 other team members from the remaining 13 players.

Number of ways to select a team under given constraints = 1 * 13C2

Number of ways to select a team without any constraints = 15C3

Probability \(= \frac{13C2}{15C3} = \frac{6}{5*7}\) or \(\frac{6*13}{35*13} = \frac{78}{455}\)



lnm87
There are 15 children in a schoolyard from which a team of 3 will be chosen. One child is named Joey; another is named Anton. What is the probability that Joey is included in the team and Anton is excluded from the team?

A. 2/15
B. 39/182
C. 1/5
D. 1/4
E. 78/455
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There are 15 children in a schoolyard from which a team of 3 will be 10 Jun 2020, 00:43
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lnm87
There are 15 children in a schoolyard from which a team of 3 will be chosen. One child is named Joey; another is named Anton. What is the probability that Joey is included in the team and Anton is excluded from the team?

A. 2/15
B. 39/182
C. 1/5
D. 1/4
E. 78/455
Joey can be selected in three ways:
J _ _ \(P(JnotA)_1\) = \(\frac{1}{15}*\frac{13}{14}*\frac{12}{13}\)
_ J _ \(P(JnotA)_2\) = \(\frac{13}{15}*\frac{1}{14}*\frac{12}{13}\)
_ _ J \(P(JnotA)_3\) = \(\frac{13}{15}*\frac{12}{14}*\frac{1}{13}\)

So, P = \(3*\frac{13*12}{15*14*13} = \frac{6*13}{35*13}\)

Answer E
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There are 15 children in a schoolyard from which a team of 3 will be 10 Jun 2020, 21:44
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lnm87
lnm87
There are 15 children in a schoolyard from which a team of 3 will be chosen. One child is named Joey; another is named Anton. What is the probability that Joey is included in the team and Anton is excluded from the team?

A. 2/15
B. 39/182
C. 1/5
D. 1/4
E. 78/455
Joey can be selected in three ways:
J _ _ \(P(JnotA)_1\) = \(\frac{1}{15}*\frac{13}{14}*\frac{12}{13}\)
_ J _ \(P(JnotA)_2\) = \(\frac{13}{15}*\frac{1}{14}*\frac{12}{13}\)
_ _ J \(P(JnotA)_3\) = \(\frac{13}{15}*\frac{12}{14}*\frac{1}{13}\)

So, P = \(3*\frac{13*12}{15*14*13} = \frac{6*13}{35*13}\)

Answer E
nick1816
Is this method acceptable?
I don't see anything wrong but just want to clarify.
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There are 15 children in a schoolyard from which a team of 3 will be 10 Jun 2020, 23:22
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Your approach is on point. Only difference in our approaches is you did the arrangements too (i only did selection). Basically, you don't have to do the arrangements in the probability questions, as it got cancelled out in the end. (might save some time, that's all)

lnm87
lnm87
lnm87
There are 15 children in a schoolyard from which a team of 3 will be chosen. One child is named Joey; another is named Anton. What is the probability that Joey is included in the team and Anton is excluded from the team?

A. 2/15
B. 39/182
C. 1/5
D. 1/4
E. 78/455
Joey can be selected in three ways:
J _ _ \(P(JnotA)_1\) = \(\frac{1}{15}*\frac{13}{14}*\frac{12}{13}\)
_ J _ \(P(JnotA)_2\) = \(\frac{13}{15}*\frac{1}{14}*\frac{12}{13}\)
_ _ J \(P(JnotA)_3\) = \(\frac{13}{15}*\frac{12}{14}*\frac{1}{13}\)

So, P = \(3*\frac{13*12}{15*14*13} = \frac{6*13}{35*13}\)

Answer E
nick1816
Is this method acceptable?
I don't see anything wrong but just want to clarify.
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There are 15 children in a schoolyard from which a team of 3 will be 10 Jun 2020, 23:47
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nick1816
Your approach is on point. Only difference in our approaches is you did the arrangements too (i only did selection). Basically, you don't have to do the arrangements in the probability questions, as it got cancelled out in the end. (might save some time, that's all)

lnm87
lnm87
lnm87
There are 15 children in a schoolyard from which a team of 3 will be chosen. One child is named Joey; another is named Anton. What is the probability that Joey is included in the team and Anton is excluded from the team?

A. 2/15
B. 39/182
C. 1/5
D. 1/4
E. 78/455
Joey can be selected in three ways:
J _ _ \(P(JnotA)_1\) = \(\frac{1}{15}*\frac{13}{14}*\frac{12}{13}\)
_ J _ \(P(JnotA)_2\) = \(\frac{13}{15}*\frac{1}{14}*\frac{12}{13}\)
_ _ J \(P(JnotA)_3\) = \(\frac{13}{15}*\frac{12}{14}*\frac{1}{13}\)

So, P = \(3*\frac{13*12}{15*14*13} = \frac{6*13}{35*13}\)

Answer E
nick1816
Is this method acceptable?
I don't see anything wrong but just want to clarify.
Thank you.
You are right about time saving as i did took time, half of which was taken by \(\frac{13}{13}\) multiplication - an additional step.
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There are 15 children in a schoolyard from which a team of 3 will be 16 Jun 2020, 17:28
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lnm87
There are 15 children in a schoolyard from which a team of 3 will be chosen. One child is named Joey; another is named Anton. What is the probability that Joey is included in the team and Anton is excluded from the team?

A. 2/15
B. 39/182
C. 1/5
D. 1/4
E. 78/455

We exclude Anton from consideration (leaving 14 children available), and since Joey is already selected, there are now only 13 children remaining to fill the two remaining vacant slots. Thus, the number of ways that Joey is included and Anton is not is:

13C2 = 13! / (2! x 11!) = (13 x 12) / 2! = 78

Without any restrictions, the number of ways to select 3 children from 15 is:

15C3 = 15! / (3! x 12!) = (15 x 14 x 13)/3! = 5 x 7 x 13 = 455

Thus, the probability is 78/455.

Answer: E
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There are 15 children in a schoolyard from which a team of 3 will be 28 Jan 2025, 07:29
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