We need to see your work to know how to help..Guessball is a team sport requiring x players on each side. Alex and I are part of 2x players drawn randomly. What's the probability of being on same side?
a) (x-1)/2x
b) 1/2
c) More info needed
d) (x-1)/(2x-1)
e) x/(2x-1)
We need to see your work to know how to help..
The choices for each of you are "random" - perhaps you could think of flipping a coin? What is the chance that two flips are the same, either HH or TT? Is that easier than you thought?
Dr. Phil and Subhotosh. Please check: I am not coming up with 1/2 as the proper answer.The way I figured it we could choose Team A x B, B x A, A x A, B x B = 1/2
Is that right?
OOPS - I fear JeffM is right! Whereas the probability that Alex will be on team A is exactly 1/2, we must use CONDITIONAL probability for which team you will be on. Since one of the positions on team A is already taken, your probability of being one of the (x-1) players on the same team, out of (2x-1) yet to be chosen, is (x-1)/(2x-1).Dr. Phil and Subhotosh. Please check: I am not coming up with 1/2 as the proper answer.
P(Alex will be chosen by team 1) = \(\displaystyle \dfrac{x}{2x} = \dfrac{1}{2}.\)
P(I will be chosen by team 1 given Alex was chosen by team 1) = \(\displaystyle \dfrac{x - 1}{2x - 1}.\)
P(Alex and I will both be on team 1) = \(\displaystyle \dfrac{x - 1}{2x - 1} * \dfrac{1}{2} = \dfrac{x - 1}{2(2x - 1)}.\)
P(Alex will be chosen by team 2) = \(\displaystyle \dfrac{x}{2x} = \dfrac{1}{2}.\)
P(I will be chosen by team 2 given Alex was chosen by team 2) = \(\displaystyle \dfrac{x - 1}{2x - 1}.\)
P(Alex and I will both be on team 2) = \(\displaystyle \dfrac{x - 1}{2x - 1} * \dfrac{1}{2} = \dfrac{x - 1}{2(2x - 1)}.\)
P(Alex and I will be on same team) = P(both on team 1) + P(both on team 2) =
\(\displaystyle \dfrac{x - 1}{2(2x - 1)} +\dfrac{x - 1}{2(2x - 1)} = \dfrac{2(x - 1)}{2(2x - 1)} = \dfrac{x - 1}{2x - 1}.\)
If I am analyzing correctly, we are not dealing with independent events.
Elegant.It's not necessary to talk divide this into "team A" and "team B". I am sure to be on a team. That leaves 2n-1 people to be chosen from and n-1 places on my team. The probability that any one of those 2n- 1 persons will be selected for one of the n-1 places on my team is \(\displaystyle \dfrac{n-1}{2n-1}\).