russ9 wrote:
There are 9 people in the room. There are two pairs of siblings within that group. If we choose two people, what is the probability that they will not be siblings?
Can someone please explain where i'm going wrong?
I'm using the approach: No Sibling Pair = 1 - Probability of sibling pair.
P = \(\frac{(9C4)(1C1)}{(9C2)}\) -- why is this wrong? Is this equation not implying that at first, I have the probability of choose 4 people out of 9, then I can only pick 1 so my top is the favorable solution. My denominator is the total number of outcomes.
The numerator is wrong. We are not choosing 4 people, we are choosing 2.
How many ways to choose a sibling pair from 9 people where there are two pairs of siblings (X, X, Y, Y, A, B, C, D, E)? Only two ways XX and YY.
Hence, P = 1 - 2/(9C2) = 1 - 1/18 = 1/17.
I am reading what you're writing but I just cant grasp it. How is this different than the other problems where we try to keep the numerator and denominator in the same units, i.e: people and not pairs. Is P = \(\frac{(9C4)(1C1)}{(9C2)}\) not saying that i want to pick ANY member out of the 4 siblings and then I can only choose 1 member after choosing that ANY member?