reply2spg 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
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. Therefore, the probability that the chosen two people are not siblings is 16/21.
Answer: E
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