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

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
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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= \(\frac{9C4+9C6}{11C6}= \frac{5}{11}\)

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
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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

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
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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

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

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 !

Hello,

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