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# Attorneys are questioning 11 potential jurors about their

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Joined: 11 Feb 2012
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30 Sep 2012, 00:14
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71% (02:22) correct 29% (02:36) wrong based on 327 sessions

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Attorneys are questioning 11 potential jurors about their ability to serve on a jury and decide a case fairly. Seven individuals will be selected to serve on the jury. If each potential juror has an equal chance of being selected, what is the probability that Tamara and Inga are chosen for the jury?

A. 2/11
B. 2/7
C. 21/55
D. 4/11
E. 34/55
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01 Oct 2012, 07:03
4
3
smartass666 wrote:
Attorneys are questioning 11 potential jurors about their ability to serve on a jury and decide a case fairly. Seven individuals will be selected to serve on the jury. If each potential juror has an equal chance of being selected, what is the probability that Tamara and Inga are chosen for the jury?

A. 2/11
B. 2/7
C. 21/55
D. 4/11
E. 34/55

Easier way would be to consider all 11 people in a row where the first 7 will be selected for the jury. We want Tamara and Inga to be among those 7: P=7/11*6/10=21/55.

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Updated on: 30 Sep 2012, 05:11
6
4
smartass666 wrote:
Attorneys are questioning 11 potential jurors about their ability to serve on a jury and decide a case fairly. Seven individuals will be selected to serve on the jury. If each potential juror has an equal chance of being selected, what is the probability that Tamara and Inga are chosen for the jury?

2/11
2/7
21/55
4/11
34/55

A direct probabilistic approach:

First person to be chosen one of the two girls - 2/11
Second person to be chosen to be the other girl - 1/10
The probability to chose both girls - (2/11)*(1/10) = 1/55.
The two girls can be chosen anywhere in the list of the 7 jurors, therefore the final probability is (1/55)*7C2 = (1/55)*(7*6/2) = 21/55.

Combinatorial approach:
Total number of possibilities to choose the 7 jurors - 11C7 = 11*10*9*8/(2*3*4) = 11*10*3.
Number of possibilities to have the two girls among the final 7 jurors - 9C5 = 9*8*7*6/(2*3*4) = 9*2*7 (the two girls must be in, therefore we need to choose 5 more people from the remaining 9).
Required probability 9*2*7/(11*10*3) = 21/55.
(I prefer not to multiply before I express the final probability as a ratio, I always try to reduce a fraction whenever possible.)

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Originally posted by EvaJager on 30 Sep 2012, 03:30.
Last edited by EvaJager on 30 Sep 2012, 05:11, edited 1 time in total.
##### General Discussion
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Joined: 11 Feb 2012
Posts: 12

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30 Sep 2012, 05:04
1
EvaJager wrote:
smartass666 wrote:
Attorneys are questioning 11 potential jurors about their ability to serve on a jury and decide a case fairly. Seven individuals will be selected to serve on the jury. If each potential juror has an equal chance of being selected, what is the probability that Tamara and Inga are chosen for the jury?

2/11
2/7
21/55
4/11
34/55

A direct probabilistic approach:
First person to be chosen one of the two girls - 2/11
Second person to be chosen to be the other girl - 1/10
The probability to chose both girls - (2/11)*(1/10) = 1/55.
The two girls can be chosen anywhere in the list of the 7 jurors, therefore the final probability is (1/55)*11C7 = (1/55)*(7*6/2) = 21/55.

Dint understand the last step : The two girls can be chosen anywhere in the list of the 7 jurors, therefore the final probability is (1/55)*11C7 = (1/55)*(7*6/2) = 21/55.

1/55 * 11C7 cannot be equal to 1/55*(7*6/2).
11C7= 330.
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30 Sep 2012, 05:10
smartass666 wrote:
EvaJager wrote:
smartass666 wrote:
Attorneys are questioning 11 potential jurors about their ability to serve on a jury and decide a case fairly. Seven individuals will be selected to serve on the jury. If each potential juror has an equal chance of being selected, what is the probability that Tamara and Inga are chosen for the jury?

2/11
2/7
21/55
4/11
34/55

A direct probabilistic approach:
First person to be chosen one of the two girls - 2/11
Second person to be chosen to be the other girl - 1/10
The probability to chose both girls - (2/11)*(1/10) = 1/55.
The two girls can be chosen anywhere in the list of the 7 jurors, therefore the final probability is (1/55)*11C7 = (1/55)*(7*6/2) = 21/55.

Dint understand the last step : The two girls can be chosen anywhere in the list of the 7 jurors, therefore the final probability is (1/55)*11C7 = (1/55)*(7*6/2) = 21/55.

1/55 * 11C7 cannot be equal to 1/55*(7*6/2).
11C7= 330.

Thank you, you are right: it should be 1/55*(7C2). The two girls can be chosen anywhere in the list of the final 7 jurors.
Corrected my original post.
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01 Oct 2012, 07:40
1
1
smartass666 wrote:
Attorneys are questioning 11 potential jurors about their ability to serve on a jury and decide a case fairly. Seven individuals will be selected to serve on the jury. If each potential juror has an equal chance of being selected, what is the probability that Tamara and Inga are chosen for the jury?

A. 2/11
B. 2/7
C. 21/55
D. 4/11
E. 34/55

Number of ways to select Tamara & Inga "for sure" amongst the 7 out of 11 jurors are: 2C2 * 9C5
Total number ways to select 7 jurors of 11 jurors are 11C7

Probability is (2C2*9C5)/11C7 = 21/55
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05 Sep 2013, 13:09
Hi,

Though i got the answer right, i initially approach did not yield the right answer.

Chance that the two girls are selected = 1- (Chance that none of the 2 girls are selected)
i.e - Ways that none of the two girls are selected = 9C7
- Total ways of selecting 7 of 11 people = 11C7

Answer = 1-(9C7/11C7) = 1-(6/55) = 49/55
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05 Sep 2013, 21:52
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1
chechaxo wrote:

Chance that the two girls are selected = 1- (Chance that none of the 2 girls are selected)

This is incorrect.

Chance that the two girls are selected = 1- (Chance that none of the 2 girls are selected + Chance that only one of the two girls is selected)

There are 3 possible cases:
1. Both the girls are not selected.
2. Only one girl is selected.
3. Both girls are selected.

It is actually easier to deal with "Both girls are selected" directly.

No of ways of selecting both girls = 9C5 (Select both girls in 1 way and then select other 5 jurors in 9C5 ways)
No of ways of selecting 7 jurors out of 11 = 11C7

Req Probability = 9C5/11C7
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28 Oct 2015, 18:25
1
I chose the combinatorics:
we need to choose 7 people:
7C11 = 55
from 7, we need to have both of them. so 2C7 = 22
probability to be both are thus 22/55
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04 Jun 2018, 08:39
VeritasPrepKarishma wrote:
chechaxo wrote:

Chance that the two girls are selected = 1- (Chance that none of the 2 girls are selected)

This is incorrect.

Chance that the two girls are selected = 1- (Chance that none of the 2 girls are selected + Chance that only one of the two girls is selected)

There are 3 possible cases:
1. Both the girls are not selected.
2. Only one girl is selected.
3. Both girls are selected.

It is actually easier to deal with "Both girls are selected" directly.

No of ways of selecting both girls = 9C5 (Select both girls in 1 way and then select other 5 jurors in 9C5 ways)
No of ways of selecting 7 jurors out of 11 = 11C7

Req Probability = 9C5/11C7

What is the probability, through combinations, of chosing only one girl?
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26 Jan 2019, 08:46
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Top Contributor
smartass666 wrote:
Attorneys are questioning 11 potential jurors about their ability to serve on a jury and decide a case fairly. Seven individuals will be selected to serve on the jury. If each potential juror has an equal chance of being selected, what is the probability that Tamara and Inga are BOTH chosen for the jury?

A. 2/11
B. 2/7
C. 21/55
D. 4/11
E. 34/55

P(Tamara and Inga are BOTH chosen for the jury) = (# of ways to choose 7 people that include Tamara and Inga)/(TOTAL # of ways to choose 7 people)

# of ways to choose 7 people that include Tamara and Inga
Stage 1: place Tamara and Inga on the jury. This can be accomplished in 1 way
Stage 2: select 5 more people from the remaining 9 people. This can be accomplished in 9C5 ways (= 126 ways)
By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus create a jury with Tamara and Inga) in (1)(126) ways (= 126 ways)

TOTAL # of ways to choose 7 people
We can choose 7 people from 11 people in 11C7 ways
11C7 = 330

So, P(Tamara and Inga are BOTH chosen for the jury) = 126/330
= 63/165
= 21/55

Cheers,
Brent
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26 Jan 2019, 09:46
1
smartass666 wrote:
Attorneys are questioning 11 potential jurors about their ability to serve on a jury and decide a case fairly. Seven individuals will be selected to serve on the jury. If each potential juror has an equal chance of being selected, what is the probability that Tamara and Inga are chosen for the jury?

A. 2/11
B. 2/7
C. 21/55
D. 4/11
E. 34/55

If Tamara and Inga are chosen for the jury, this means that remaining jurors can be chosen in$$9C_5$$ ways

Number of ways of selecting 7 jurors from 11 potential jurors = $$11C_7$$ ways

Probability = 9C5 / 11C7 = 21/55

C
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27 Jan 2019, 06:56
1
KanishkM wrote:
If Tamara and Inga are chosen for the jury, this means that remaining jurors can be chosen in$$9C_5$$ ways

Exactly KanishkM !

I explain to my students that they should think Tamara and Inga "took" two places in advance, so that we have in fact 7-2 = 5 places available, to be occupied from people chosen among 11-2 individuals.

Regards,
Fabio.
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Re: Attorneys are questioning 11 potential jurors about their   [#permalink] 27 Jan 2019, 06:56
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