Hi viju,
Among the seven people, we have two pairs and a trio. In each of the pairs, the two people are siblings to each other--that's 4 people each of whom have exactly 1 sibling. In the trio, each person is sibling to the other two--that's 3 people, each of whom have exactly 2 siblings.
EDIT: And, just to elaborate on papgust's solution, whenever a probability problem says "at least" it is almost always easier to think about the undesired outcomes (what you don't want) than the desired outcomes, and so use the formula:
Prob total = 1 = prob desired + prob undesired or
prob desired = 1 - prob undesired (this is the approach papgust used)
Again, solve for undesired first and subtract from total.
Similarly, when dealing with combinatorics, if there is a restriction imposed use:
Total number of selections = permitted number + restricted number (we can also use this approach here)
Again, in combinatorics, where there is a restriction imposed and if the question is asking you to solve for permitted, it is almost always easiest to solve for total and restricted, and then to subtract restricted from total to compute permitted.
Here, we are asked for the probability of NOT getting siblings. So, after figuring out the total number of ways (7C2), it will be easier to think about the ways we CAN get the siblings, and then subtract this from total. As papgust and I wrote above, we can deduce that there are two different pairs, and then a trio of siblings. How can we select two siblings then? Either by selecting either of the two pairs--that's 2 ways...OR by selecting any 2 siblings from the trio, which is 3C2, or another 3 ways. So there are 5 ways we CAN select siblings. There are 21 ways to select 2 people in total. So, there are 21 - 5= 16 (out of 21) selections of 2 people that do NOT have siblings.
Last edited by
Testluv on Wed Dec 16, 2009 11:40 pm, edited 2 times in total.
Kaplan Teacher in Toronto
