Answer:
Follows are the solution to the given point:
Step-by-step explanation:
For option A:
In the first point let z be the number of heads which is available on the first two trails of tosses so, the equation is:
[tex]P(X=k | z=2 ) = \begin{pmatrix} 10-2\\ k-2\end{pmatrix} (\frac{1}{2})^{k-2} (\frac{1}{2})^{10-k}[/tex]
[tex]= \begin{pmatrix} 8\\k-2\end{pmatrix} (\frac{1}{2})^{k-2} (\frac{1}{2})^{10-k} \ \ \ \ \ \ \ \ \ \ \\\\[/tex]
[tex]k= 2,3..........10[/tex]For option B:
[tex]P(X=k | X \geq 2 ) = \sum^{10}_{i=2} \begin{pmatrix} 10-i\\ k-i\end{pmatrix} (\frac{1}{2})^{k-i} (\frac{1}{2})^{10-k}[/tex]
[tex]k= 2,3, 4.........10[/tex]
