Attorneys are questioning 11 potential jurors about their

Question

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

EvaJager · Accepted Answer

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.)

Answer C.

KarishmaB · Answer

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

smartass666 · Answer

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.

EvaJager · Answer

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.

Bunuel · Answer

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.

Answer: C.

navigator123 · Answer

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

chechaxo · Answer

Hi,

Though i got the answer right, i initially approach did not yield the right answer.
Please help me identify the error in my thought process

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

mvictor · Answer

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

alonsotruj · Answer

What is the probability, through combinations, of chosing only one girl?

BrentGMATPrepNow · Answer

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

Answer: C

Cheers,
Brent

KanishkM · Answer

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

fskilnik · Answer

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.

Rishm · Answer

So, in this question, there are total 7 spots to be filled by 11 people, use combinations - which gives us 11!/7!*4! = 330 (I got this as total), then for fav outcomes, take both girls as 1 in the 7 slots, you are left with 5 slots now. Now choose 9 people with those 5 slots, same drill combinations 9!/4!5!= 126. Ensure you dont put 11 as people as both these girls are out. Now 126/330 gives us 21/55. I trust this should be a 685 question.

luisdicampo · Answer

Deconstructing the Question

We are selecting 7 jurors from 11 people. We want the probability that both Tamara and Inga are selected.

Step-by-step

Total number of ways to choose 7 jurors:

\binom{11}{7}

Now count the favorable outcomes.

If Tamara and Inga are included, we must choose the remaining 5 jurors from the other 9 people:

\binom{9}{5}

So the probability is:

\frac{\binom{9}{5}}{\binom{11}{7}}

Compute:

\binom{9}{5} = 126

\binom{11}{7} = 330

\frac{126}{330} = \frac{21}{55}

Answer C

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