
David B. answered 05/19/21
Math and Statistics need not be scary
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