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 individuals 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.
Please, can anyone assist me with this PS question? I'm confused with it. Thanks in advance!
In a room filled with 7 people, 4 people phave exactly 1
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First we need to recognize that the given information tells us that the 7 people consist of:BTGmoderatorLU 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 individuals 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
- a sibling trio
- a sibling pair
- and another sibling pair
Using counting techniques:
For this question, it's easier to find the complement.
So P(not siblings) = 1 - P(they are siblings)
P(they are siblings) = [# of ways to select 2 siblings] / [total # of ways to select 2 people]
# of ways to select 2 siblings
Case a) 2 siblings from the sibling trio: from these 3 siblings, we can select 2 siblings in 3C2 ways (3 ways)
Case b) 2 siblings from first sibling pair: we can select 2 siblings in 2C2 ways (1 way)
Case c) 2 siblings from second sibling pair: we can select 2 siblings in 2C2 ways (1 way)
So, total number of ways to select 2 siblings = 3+1+1 = 5
total # of ways to select 2 people
We have 7 people and we want to select 2 of them
We can accomplish this in 7C2 ways (21 ways)
So, P(they are siblings) = 5/21
This means P(not siblings) = 1 - 5/21
= 16/21
Answer: E
Cheers,
Brent
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Hi All,
Brent's approach focuses on the use of the Combination Formula, which is probably the most "elegant" way to solve this problem. There are other ways to solve this problem though, if you're not yet comfortable with doing that type of math. Here's an option that focuses on straight "probability math"....
The first step is to organize the 7 people into family-groups. 4 people have exactly 1 sibling. We'll call those people:
A1, A2
B1, B2
A1 and A2 are siblings, etc.
3 people have exactly 2 siblings. We'll call them:
C1, C2 and C3
We're going to select 2 people from the room; what is the probability that they are NOT siblings.
Probability = (# of ways you want something to happen)/(# of total possibilities)
The total number of possibilities = (7)(6) = 42
Now let's figure out the ways that give us 2 non-siblings:
If the first person selected is:
A1, then there are 5 non-sibling options
A2, then there are 5 non-sibling options
B1, then there are 5 non-sibling options
B2, then there are 5 non-sibling options
C1, then there are 4 non-sibling options
C2, then there are 4 non-sibling options
C3, then there are 4 non-sibling options
The total number of 2 non-sibling options is 5+5+5+5+4+4+4 = 32
The probability is 32/42 = 16/21
Final Answer: E
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Rich
Brent's approach focuses on the use of the Combination Formula, which is probably the most "elegant" way to solve this problem. There are other ways to solve this problem though, if you're not yet comfortable with doing that type of math. Here's an option that focuses on straight "probability math"....
The first step is to organize the 7 people into family-groups. 4 people have exactly 1 sibling. We'll call those people:
A1, A2
B1, B2
A1 and A2 are siblings, etc.
3 people have exactly 2 siblings. We'll call them:
C1, C2 and C3
We're going to select 2 people from the room; what is the probability that they are NOT siblings.
Probability = (# of ways you want something to happen)/(# of total possibilities)
The total number of possibilities = (7)(6) = 42
Now let's figure out the ways that give us 2 non-siblings:
If the first person selected is:
A1, then there are 5 non-sibling options
A2, then there are 5 non-sibling options
B1, then there are 5 non-sibling options
B2, then there are 5 non-sibling options
C1, then there are 4 non-sibling options
C2, then there are 4 non-sibling options
C3, then there are 4 non-sibling options
The total number of 2 non-sibling options is 5+5+5+5+4+4+4 = 32
The probability is 32/42 = 16/21
Final Answer: E
GMAT assassins aren't born, they're made,
Rich
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Let's say that the 7 people are ABCDEFG.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 individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?
5/21
3/7
4/7
5/7
16/21
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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That is one way, I think it is easier to calculate the probabilities than the combinations.
there is 4/7 chance of selecting a person with one sibling initially with a 5/6 probability the second person will not be a sibling.
There is a 3/7 Chance of selecting a person with 2 siblings initially with 4/6 probability the second person will not be a sibling.
Simply adding these probabilities together gives you the answer.
4/7 * 5/6 + 3/7 * 4/6 = 20/42 + 12/42 = 32/42 = 16/21.
Regards!
there is 4/7 chance of selecting a person with one sibling initially with a 5/6 probability the second person will not be a sibling.
There is a 3/7 Chance of selecting a person with 2 siblings initially with 4/6 probability the second person will not be a sibling.
Simply adding these probabilities together gives you the answer.
4/7 * 5/6 + 3/7 * 4/6 = 20/42 + 12/42 = 32/42 = 16/21.
Regards!