In a room filled with 7 people, 4 people have exactly 1....

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swerve: Legendary Member; Posts: 2499; Joined: Sun Oct 29, 2017 2:04 pm; Followed by:6 members

In a room filled with 7 people, 4 people have exactly 1....

by swerve » Tue Dec 19, 2017 10:33 am

In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two indivuduals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

A) 5/21
B) 3/7
C) 4/7
D) 5/7
E) 16/21

The OA is E.

I am unable to interpret statements

1) 4 people have exactly 1 sibling in the room
2) 3 people have exactly 2 siblings in the room
3) no of ways people siblings can be selected

Experts, can you help me with this PS question, please? Thanks!

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Source: Beat The GMAT — Problem Solving | Tue Dec 19, 2017 10:33 am

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by GMATGuruNY » Tue Dec 19, 2017 10:40 am

swerve wrote:In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two indivuduals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

A) 5/21
B) 3/7
C) 4/7
D) 5/7
E) 16/21

Let's say that the 7 people are ABCDEFG.

4 people have exactly 1 sibling:
Let's say that A and B are siblings and that C and D are siblings.
This means:
A has 1 sibling (B).
B has 1 sibling (A).
C has 1 sibling (D).
D has 1 sibling (C).

3 people have exactly 2 siblings:
Let's say that E, F and G are all siblings of each other.
This means:
E has 2 siblings (F and G).
F has 2 siblings (E and G).
G has 2 siblings (E and F).

Total number of sibling pairs = 5: AB, CD, EF, EG, FG.
Total number of pairs that can be formed from 7 people: 7C2 = 21.
P(sibling pair) = 5/21
P(not sibling pair) = 1 - 5/21 = 16/21.

The correct answer is E.

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regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

by regor60 » Tue Dec 19, 2017 11:26 am

swerve wrote:In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two indivuduals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

A) 5/21
B) 3/7
C) 4/7
D) 5/7
E) 16/21

The OA is E.

I am unable to interpret statements

1) 4 people have exactly 1 sibling in the room
2) 3 people have exactly 2 siblings in the room
3) no of ways people siblings can be selected

Experts, can you help me with this PS question, please? Thanks!

Another way is via probability.

If 4 people each have one sibling, then that means there are two distinct sibling pairs.
Likewise, there are 3 people with two siblings. The group of 3 are each other's siblings.

Label the first sibling pair A1 and A2, Second B1 and B2. And the 3 people C1, C2 and C3.

Pick one person at random. The probability it's from the A group is 2/7. The probability of NOT picking the other sibling in A is 5/6. Together, the probability is 2/7*5/6 or 10/42.

Likewise, the probability the first person was chosen from the B group is also 2/7 and the probability of NOT picking again from B is 5/6, for a total of 10/42 again.

Finally, the probability of picking first from the C group is 3/7. The probability of picking NOT from C on the second pick is 4/6. Total probability for C is 3/7*4/6 = 12/42.

Adding the probabilities 10/42 + 10/42 + 12/42 = 32/42 = [spoiler]16/21, E[/spoiler]

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Thu Sep 12, 2019 6:23 pm

swerve wrote:In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two indivuduals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

A) 5/21
B) 3/7
C) 4/7
D) 5/7
E) 16/21

Let A, B, C, D, E, F and G be the 7 people in the room. To satisfy the condition that 4 people have exactly 1 sibling and 3 people have exactly 2 siblings, we can let A and B be siblings (but not to other people), C and D be siblings (but not to other people), and E, F and G are siblings (but not to other people).

Let's consider the probability of how each person is chosen:

If A is chosen first, then B can't be chosen. So the probability is:

1/7 x 5/6 = 5/42

This probability will be the same if B, C, or D is chosen first.

If E is chosen first, then neither F nor G can be chosen. So the probability is:

1/7 x 4/6 = 4/42

This probability will be the same if F or G is chosen first.

Therefore, the overall probability is:

5/42 x 4 + 4/42 x 3 = 20/42 + 12/42 = 32/42 = 16/21

Alternate Solution:

Notice that 2 people can be chosen out of 7 people in 7C2 = 7!/(5!*2!) = (7 x 6)/2 = 21 ways.

With A, B, C, D, E, F, and G as above, we see that there are 5 ways to choose a sibling pair: A-B, C-D, E-F, E-G and F-G. Thus, 21 - 5 = 16 choices of do not include a sibling pair. Thus, the probability that the chosen two people are not siblings is 16/21.

Answer: E

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