fibbonnaci wrote:this is a simple one provided u get the logic rite.
lets say A, B, C, D are the 4 ppl who have one sibling. and E, F, and G are the people who have 2 siblings.
eg say A has a sibling within the 4 people and E has a sibling within the three people.
so in order for selecting 2 people in such a way that they are not siblings, we must select people from different groups.
selecting one person from the 4 membered group is 4C1= 4
selecting one person from the 3 membered group is 3C1 = 3
number of ways= 12 (4*3)
total number of possibilities of selecting 2 people out of the 7 membered group= 7C2= 21
probability= 12/21 => 3/7
the probability of selecting 2 individuals such that they are not siblings is 3/7
Hope this helps!
Not quite!
You've assumed that you have to select one person from each group. However, in the first group of 4, each person only has 1 sibling, i.e. there are 2 pairs of siblings in that set of 4 people. So, there are some pairs you can pick out of that group of 4 who won't be siblings.
We could use your approach, but we have to break it down further:
Siblings: AB, CD, EFG
We can pick one person from any two of these three groups and end up with a non-sibling pair.
Case 1: 1 from AB, 1 from CD: 2*2 = 4 pairs
Case 2: 1 from AB, 1 from EFG: 2*3 = 6 pairs
Case 3: 1 from CD, 1 from EFG: 2*3 = 6 pairs
So, we can make 16 non-sibling pairs.
There are 7C2 = 21 total pairs available.
Accordingly, the probability of a non-sibling pair is 16/21.
