A and B are members of a group of 11 people. A team of 6 members is to

Question

A and B are members of a group of 11 people. A team of 6 members is to be chosen from this group. What is the probability that A and B will either both be in the team or will both be out of the team?

A. 3/11
B. 54/121
C. 5/11
D. 6/11
E. 9/11

globaldesi · Accepted Answer

when both A and B are included (A and B already there. select 4 from remaining
ways =  \frac{9C_4}{11C_6}

when neither is included 
select 6 from 9
ways =  \frac{9C_6}{11C_6} 
Total ways =  \frac{9C_4}{11C_6}  +  \frac{9C_6}{11C_6} 
solve and value will be  \frac{5}{11} 
Thus C

nick1816 · Answer

Total number of ways to select a team when both A and B are in the team= 9C4

Total number of ways to select a team when both A and B are out of the team= 9C6

Probability=  \frac{9C4+9C6}{11C6}= \frac{5}{11}

chetan2u · Answer

Many ways to do it..

(I) As also shown above
Ways to select both A and B in the team= 9C4= \frac{9*8*7*6*5!}{4!*5!}=\frac{9*8*7*6}{4*3*2}=126 
Ways in which both A and B are out of the team= 9C6= \frac{9*8*7*6!}{3!*6!}=\frac{9*8*7}{3*2}=84 
Total ways =11C6= \frac{11*10*9*8*7*6!}{6!*5!}=\frac{11*10*9*8*7}{5*4*3*2}=462 
Probability =  \frac{9C4+9C6}{11C6}=\frac{210}{462}=\frac{5}{11}

(II) Take ways in which ONLY one is there, so 1 seat is for either of two, so two ways
Rest 5 seats by remaining 9 in 9C5 =126
Total = 2*126=252
Total ways without restriction = 11C6=462
Probability  \frac{252}{462}=\frac{6}{11} 
Our answer =  1-\frac{6}{11}=\frac{5}{11}

C

abhishek001 · Answer

Answer is C.

Two cases 
1.with both A and B
4 to be selected out of 9 as a and b are already selected
So 9C4.

2. Neither A nor B
So 6 people to be selected out of 9 as A and B are already discarded.
So 9C6

Total outcomes
11C6
Therefore 9C4+9C6/11C6=5/11.

Posted from my mobile device

ScottTargetTestPrep · Answer

The probability that both A and B are on the team is 9C4/11C6:

9C4 = 9! / (4! x 5!) = (9 x 8 x 7 x 6) / (4 x 3 x 2) = 9 x 2 x 7 = 126

11C6 = 11! / (6! x 5!) = (11 x 10 x 9 x 8 x 7 x 6) / (6 x 5 x 4 x 3 x 2) = 11 x 2x 3 x 7 = 462
 
The probability that both A and B are not on the team is 9C6/11C6:

9C6 = 9! / (6! x 3!) = (9 x 8 x 7 x 6 x 5 x 4) (6 x 5 x 4 x 3 x 2) = 3 x 4 x 7 = 84

11C6 = 462

Thus, the overall probability is:

126/420 + 84/429= 210/462 = 70/154 = 10/22 = 5/11.

Answer: C

gmatapprentice · Answer

Bunuel   VeritasKarishma   chetan2u

The tough part in this question is solving the probability - any tips on how this can be done quickly. Thanks.

(9C4+9C6)/11C6

KarishmaB · Answer

I would go "Finding the complementary probability" way because that involves one calculation.

P (Only one of A and B in the team) = 2 * P(Only A in the team) = 2 * 9C5/11C6

9C5 is the same as 9C4 and 11C6 is the same as 11C5.

9C4 = 9*8*7*6/4*3*2*1 = 9 * 2 * 7

11C5 = 11*10*9*8*7 / 5*4*3*2*1 = 11 * 3 * 2 * 7

9C4/11C5 = 3/11

P (Only one of A and B in the team) = 2 * 3/11 = 6/11

Required probability = 1 - 6/11 = 5/11

Answer (C)

gmatapprentice · Answer

VeritasKarishma  thanks didnt realize that  P (Only one of A and B in the team) = 2 * P(Only A in the team) ..good tip!

Ninua · Answer

Hello everyone,

I used the following method and I got the right answer. Is this method universally functional for this type of question or did I just get lucky ?

- First I calculated the probability of A and B being teammates of the chosen group of 6 people : (6/11) * (5/10) = (3/11)
- Then I calculated the probability of A and B being together in the group of 'the non chosen' which is 5 people out of 11 : (5/11) * (4/10) = (2/11)
- If we sum both probabilities we get (3/11) + (2/11) = (5/11) which happens to be the right answer

So is the reasoning correct or was it a lucky coincidence ?

Thank you for your attention and have a great day !

Engineer1 · Answer

Hello,

I do not understand the 9C4 part. A & B both in the team, why cannot this be 6C2, instead of 9C4?

snowinmay · Answer

Why not just use probability instead of combinations?
P(both in) = 6/11 *5/10 = 3/11
P(both out) = 5/11 * 4/10 = 2/11
Add together = 5/11

egmat · Answer

Hi snowinmay,

Good instinct, and the short answer is:  your method is completely valid, not a coincidence.  It's the same probability, just counted one person at a time instead of in a single combination.

Here's why it works. "Choose a team of 6 from 11" is the same as lining people up and asking about the first 6 slots - the probability that any specific person lands in the team is  6/11 , and order doesn't change the count as long as you're consistent.

Both A and B in: 
- P(A is in) =  6/11 
- Given A took one of the 6 spots,  5  team spots remain out of  10  remaining people, so P(B also in) =  5/10 
- Multiply:  6/11  x  5/10  =  3/11

Both A and B out: 
- P(A is out) =  5/11  (5 of the 11 are non-team spots)
- Given A is out,  4  non-team spots remain out of  10 , so P(B also out) =  4/10 
- Multiply:  5/11  x  4/10  =  2/11

The two cases can't both happen, so you add:  3/11  +  2/11  =  5/11 .

The combinations method ( 9C4  +  9C6 )/ 11C6  and your sequential method are just two languages for the same count - both correct. Use whichever is faster for you; here, yours is.

One caution to keep it safe:  the conditional fractions must update correctly each step (after placing A, both the team spots  and  the people left drop by one). You did that right. As long as you track those denominators honestly, sequential probability works for any "both in / both out" selection problem.

Answer: C

A and B are members of a group of 11 people. A team of 6 members is to

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

A and B are members of a group of 11 people. A team of 6 members is to

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards