Chapter 4 - Discrete Random Variables - Exercises 4.50 - 4.76 - Applying the Concepts - Basic - Page 206: 4.64d

$p(x\leq1)=.6386$

$p(x=1)=\begin{pmatrix} 5 \\ 1 \end{pmatrix}(.25)^1(.75)^{5-12}=\frac{5!}{1!(5-1)!}(.25)^12(.75)^4=5(.25)^1(.75)^4=.3955$ $p(x=0)=\begin{pmatrix} 5 \\ 0 \end{pmatrix}(.25)^0(.75)^{5-0}=\frac{5!}{0!(5-0)!}(.25)^0(.75)^5=1(.75)^5=.2373$ $p(x\leq1)=p(x=0)+p(x=1)=.3995+.2373=.6386$