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
A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21
It is first important to understand the problem. So let us first assume a specific case.
1. Assume the 7 people are A,B, C, D, E, F and G
2. Assume the 4 people who have 1 sibling are A, B, C and D
3. Let's assume A's sibling is B. Therefore B's sibling is A. Similarly for C and D.
4. So we are left with E, F and G. Each should have exactly 2 siblings.
5. E's siblings will be F and G. So F's siblings will be E and G and G's siblings will be E and F
6. Now look at the general case.
7.Total number of ways of selecting 2 people out of 7 people is 7C2=21
8. Instead of A, B, C and D assume any 4 people. We can see for every such 4 people assumed, there are 2 cases where the selected 2 will be siblings. In the case we assumed they are A and B or C and D. This gives one of the favorable outcomes
9. Or the 2 people selected being siblings may come out of the 3 siblings. The number of favorable outcomes is 3 as we can see in the specific case they are E and F, or F and G or E and G.
10. The total number of favorable outcomes for the selected two being siblings is 2+3=5.
11. The probability that the two selected are siblings is 5/21.
12, Therefore the probability that the two selected are not siblings is 1-5/21= 16/21