Inactive Tutor answered 05/19/21
First off, we determine how many tosses we are going to make. And it isn't hard.
We have only two outcomes, head or tail. If we have to stop when 2 head or 2 tails appear we have only six outcomes , two with two tosses and two with three tosses. (order matters because we stop only after the conditions are met. )
1. H. H. (X = two tosses). P = .16
2 T. T. (X = two tosses). P = .36
3. T. H. H (X = three tosses). P = .6x.4x.4 = .096
4 T. H. T. (X = three tosses) P = .144
5. H. T. H. (X = three tosses) P = .096
6. H. T. T. (X = three tosses) P = .144
Expected value is. ∑ P(Xi)·Xi = 3*2*.144 + 3*2*.096 + 2*.36 + 2*.16 = 2.48
easy as 3.1415926