Answer:
Step-by-step explanation:
From the given information, it is likely that the random variable(X) have the values below:
Let head be H
Let tail be T
So;
X(HH) = 2;
X(HT) = 1;
X(TH) = 1;
X(TT) = 0
The distribution can now be computed as:
[tex]p(X= TT) = \dfrac{1}{4}[/tex]
[tex]p(X=TH) = \dfrac{1}{4}[/tex]
[tex]p(X=HT) = \dfrac{1}{4}[/tex]
[tex]p(X=HH)= \dfrac{1}{4}[/tex]
Now, the expected value that is equivalent to the number of heads when the coin is flipped twice is:
[tex]E(X) = p(TT)*X(TT)+p(TH)*X(TH)+p(HT)*X(HT)+p(HH)*X(HH)[/tex]
[tex]E(X) = \dfrac{1}{4}\times 0 + \dfrac{1}{4}\times 1 + \dfrac{1}{4}\times 1 + \dfrac{1}{4}\times 2[/tex]
[tex]E(X) = 0 + \dfrac{1}{4}+ \dfrac{1}{4} + \dfrac{1}{2}[/tex]
[tex]E(X) =\dfrac{1+1+2}{4}[/tex]
[tex]E(X) =\dfrac{4}{4}[/tex]
E(X) = 1
[tex]E(X^2) = p(TT)*X(TT)^2+p(TH)*X(TH)^2+p(HT)*X(HT)^2+p(HH)*X(HH)^2[/tex]
[tex]E(X^2) = \dfrac{1}{4}\times 0^2+ \dfrac{1}{4}\times 1^2 + \dfrac{1}{4}\times 1^2 + \dfrac{1}{4}\times 2^2[/tex]
[tex]E(X^2) = 0 + \dfrac{1}{4}+ \dfrac{1}{4} + \dfrac{4}{4}[/tex]
[tex]E(X^2) =\dfrac{1+1+4}{4}[/tex]
[tex]E(X^2) =\dfrac{6}{4}[/tex]
[tex]E(X^2) =1.5[/tex]
Finally; To compute E²[X]
E²[X] = E[X]²
E²[X] = 1²
E²[X] = 1
