159. The device consists of three independently operating elements. The probabilities of faultless operation of the elements (over time $t$) are respectively: $p_{1}=0.7 ; p_{2}=0.8$; $p_{3}=0.9$. Find the probabilities that over time $t$ the elements will operate faultlessly: a) all elements; b) two elements; c) one element; d) none of the elements.
The problem is solved. The probabilities of the elements working without failure are: $p_{1}=0.7 ; p_{2}=0.8 ; p_{3}=0.9$, therefore the probabilities that the elements will fail are $q_{1}=0.3 ; q_{2}=0.2 ; q_{3}=0.1$.

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Let's construct the generating function:

$$
\begin{aligned}
& \varphi_{3}(z)=\left(p_{1} z+q_{1}\right)\left(p_{2} z+q_{2}\right)\left(p_{3} z+q_{3}\right)= \\
& =(0.7 z+0.3)(0.8 z+0.2)(0.9 z+0.1)= \\
& =0.504 z^{3}+0.398 z^{2}+0.092 z+0.006 .
\end{aligned}
$$

a) The probability that all three elements will work without failure is the coefficient of $z^{3}: P_{3}(3)=0.504$.

b) The probability that two elements will work without failure is the coefficient of $z^{2}: P_{s}(2)=0.398$.

c) The probability that one element will work without failure is the coefficient of $z^{1}: P_{3}(1)=0.092$.

d) The probability that none of the elements will work without failure is the constant term: $P_{3}(0)=0.006$.

Verification: $0.504+0.398+0.092+0.006=1$.