Chapter 3 - Section 3-2 - Probability Distributions and Probability Mass Functions - Exercises - Page 70: 3-37

$f(x)=\left\{\begin{array}{ll}{0.72,} & {x=0} \\ {0.26,} & {x=1} \\ {0.02,} & {x=2} \end{array}\right.$

P( first device functioning)=0.8, P( second device functioning)=0.9, X∈{0,1,2}, because X denotes the number of failed cases, $P(X=0)=0.8\times0.9=0.72$ $P(X=1)=0.8\times0.1+0.2\times0.9=0.26$ $P(X=2)=0.1\times0.2=0.02$