Probability/ combination

[leonswati]

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.......

[knight247]

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

[smartass666]

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?

[Ian Stewart]

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.

[LalaB]

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

[Brent@GMATPrepNow]

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

[fskilnik@GMATH]

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.