Four people have exactly one sibling and three people have exactly two siblings.
Think about how this can be done...the only way to organize is this is to have two pairs of siblings and one trio of siblings. In other words:
AB (one sibling pair)
CD (another sibling pair)
EFG (a trio of siblings)
This way each of A and B are siblings, each of C and D are siblings, so we have four people each with exactly one sibling. And each of E, F, and G are siblings so we have three people each of whom has exactly two siblings.
Because this is a "NOT" probability question, we should see how many ways we CAN select two individuals who are siblings.
Well, we can select the AB pair or the CD pair. So far, that's 2 ways. We can also select any two people from the EFG trio, so that's another 3C2 or 3 ways.
So, there are a total of 2+3 = 5 ways of pulling out 2 individuals who are siblings.
Probability = (#desired)/(#total)
The denominator of the formula is just all the ways we can select any two people from the seven. So we have:
Probability of selecting 2 people who are siblings = 5/7C2 = 5/21
Therefore, the probability of selecting 2 people who are NOT siblings is:
1 - 5/21 = 16/21
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