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!
In a room filled with 7 people, 4 people have exactly 1....
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Let's say that the 7 people are ABCDEFG.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
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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Another way is via probability.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!
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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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).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 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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