Flip a fair coin twice:
eg. if the first flip is “head” and the second is “tails” we represent this by ω = (H,T).
So Ω = {(H,H),(H,T),(T,H),(T,T)} has 4 elements.
We’ll assign prob’s to all subsets of Ω. So F consists of all
16 subsets of Ω.
eg. the event that the flips agree is {(H,H),(T,T)}, and
contains 2 outcomes.
“Fair” means all 4 outcomes are equally likely.
So let P(A) = #(A)/4.
eg. prob of 2 heads is P({(H,H)}) = 1. 4
eg. prob that flips agree is P({(H,H),(T,T)}) = 1. 2
Let the random variable X be the number of heads.
X((H,H))=2
X((H,T)) = 1 = X((T,H))
X((T,T)) = 0.
So P({ω | X(ω) = 2}) = 1/4, P({ω | X(ω) = 1}) = 1/2
I understand everything up to this point. I am confused of how the rest came to be.
There are 16 events in F:
1 with 0 outcomes, namely ∅
4 with 1 outcome, eg {(H,H)} (both flips are heads)
6 with 2 outcomes, eg {(H,H),(H,T)} (first flip is a head)
4 with 3 outcomes, eg {(H,H),(H,T),(T,H)} (at least one
is a head)
1 with 4 outcomes, namely Ω itself.
I was wondering if someone could please explain to me.
Best Answer
From the question, we have the set $\Omega = \{(H,H),(H,T),(T,H),(T,T)\}.$ In addition, the question has given the name $\mathcal F$ to the set of all subsets of $\Omega.$ Another way to say this is that $\mathcal F$ is the power set of $\Omega.$
As mentioned in the question, since $\Omega$ has four elements, its power set has $2^4 = 16$ elements. Here they are:
\begin{gather} \emptyset\text{ (the empty set, also written $\{\}$)},\\ \{(H,H)\}, \\ \{(H,T)\}, \\ \{(T,H)\}, \\ \{(T,T)\}, \\ \{(H,H),(H,T)\}, \\ \{(H,H),(T,H)\}, \\ \{(H,H),(T,T)\}, \\ \{(H,T),(T,H)\}, \\ \{(H,T),(T,T)\}, \\ \{(T,H),(T,T)\}, \\ \{(H,H),(H,T),(T,H)\} ,\\ \{(H,H),(H,T),(T,T)\}, \\ \{(H,H),(T,H),(T,T)\}, \\ \{(H,T),(T,H),(T,T)\}, \\ \{(H,H),(H,T),(T,H),(T,T)\}. \end{gather}
You may want to verify for yourself that every subset of $\Omega$ is shown on one of the lines above and that no subset is shown more than once. Then you can count for yourself how many subsets there are in the list that have zero elements, how many have exactly one element, how many have exactly two elements, and so forth.