Probability/ combination

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leonswati: Senior | Next Rank: 100 Posts; Posts: 94; Joined: Sun Aug 29, 2010 3:40 amFollowed by: 1 member; Thanked: 3 times

by leonswati » Thu Sep 15, 2011 12:00 pm

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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?
2/11

2/7

21/55

4/11

34/55

I got the answer as 4/11 which is a wrong answer. However this is how I got 4/11:

Total Combination : 11C7 = 330

Combinations where the two will be together : 10C7 = 120

So probability where the two will be together is = 120/330

= 4/11

Plz help me get the right answer.......

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Source: Beat The GMAT — Problem Solving | Thu Sep 15, 2011 12:00 pm

knight247: Legendary Member; Posts: 504; Joined: Tue Apr 19, 2011 1:40 pmFollowed by: 11 members; Thanked: 114 times

by knight247 » Thu Sep 15, 2011 12:53 pm

This topic tends to be a little annoying at the start. After rigorous practice it becomes relatively smooth.

Your 1st part is correct. 11C7=330
Which is the number of ways to select 7 ppl out of the 11 potential candidates.

Now, the number of selections where both will be included...Well, lets split that up in two parts
Out of the two, we need both to be selected which can happen in 2C2 ways. (2C2=1 but included that here so it would be easy to understand the concept)

Now, there are 5 seats still left and 9 ppl to choose from. So we select these 5 ppl from the 9 in
9C5 ways. Which is 165.

Combining both we get 2C2*9C5=165

165/330=21/55. Hence C

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smartass666: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Thu Feb 09, 2012 8:23 pm

by smartass666 » Sat Sep 29, 2012 12:57 pm

knight247 wrote:This topic tends to be a little annoying at the start. After rigorous practice it becomes relatively smooth.

Your 1st part is correct. 11C7=330
Which is the number of ways to select 7 ppl out of the 11 potential candidates.

Now, the number of selections where both will be included...Well, lets split that up in two parts
Out of the two, we need both to be selected which can happen in 2C2 ways. (2C2=1 but included that here so it would be easy to understand the concept)

Now, there are 5 seats still left and 9 ppl to choose from. So we select these 5 ppl from the 9 in
9C5 ways. Which is 165.

Combining both we get 2C2*9C5=165

165/330=21/55. Hence C

9C5 = 126..How did you get that answer?

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

by Ian Stewart » Sat Sep 29, 2012 1:18 pm

Or you can just imagine lining 11 people up at random, and choosing the first 7 for the jury. The probability Tamara is in one of the first 7 spots in line is 7/11. If Tamara is in one of those spots, the probability Inga is in one of the 6 remaining spots at the front of the line is then 6/10. Since we need both of these things to happen, we multiply these probabilities to get the answer: (7/11)(6/10) = 21/55.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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LalaB: Master | Next Rank: 500 Posts; Posts: 425; Joined: Wed Dec 08, 2010 9:00 amFollowed by: 7 members; Thanked: 56 times; GMAT Score:690

by LalaB » Mon Oct 01, 2012 8:41 am

2C2*9C5/11C7=21/55

Happy are those who dream dreams and are ready to pay the price to make them come true.(c)

In order to succeed, your desire for success should be greater than your fear of failure.(c)

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Sat Jan 26, 2019 7:36 am

leonswati 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

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

Brent Hanneson - Creator of GMATPrepNow.com

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Sun Jan 27, 2019 5:39 am

leonswati 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

$$?\,\, = \,\,{{C\left( {11 - 2\,,\,7 - 2} \right)} \over {C\left( {11,7} \right)}}\,\,\mathop = \limits^{\left( * \right)} \,\,{{9 \cdot 7 \cdot 2} \over {11 \cdot 10 \cdot 3}}\,\, = \,\,{{3 \cdot 7} \over {11 \cdot 5}}\,\, = {{21} \over {55}}$$
$$\left( * \right)\,\,\,\left\{ \matrix{
\,C\left( {9,5} \right) = {{9 \cdot 8 \cdot 7 \cdot 6} \over {4 \cdot 3 \cdot 2}} = 9 \cdot 7 \cdot 2 \hfill \cr
\,C\left( {11,7} \right) = {{11 \cdot 10 \cdot 9 \cdot 8} \over {4 \cdot 3 \cdot 2}} = 11 \cdot 10 \cdot 3 \hfill \cr} \right.$$

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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