Probability 80560

I have 3 sources, and their failure probability is 0.1. Calculate the probability that:
a) none will have a malfunction
b) 1 will have a breakdown
c) at least 1 will have a fault
d) they will all have a breakdown

Correct answer:

a =  0.729
b =  0.243
c =  0.271
d =  0.001

Step-by-step explanation:

$$\begin{gathered} p=0.1 \\ n=3 \\ a=\left(\genfrac{}{}{0ex}{}{n}{0}\right)\cdot p^0\cdot (1-p{)}^{n-0}=(\genfrac{}{}{0ex}{}{3}{0})\cdot 0.1^0\cdot (1-0.1{)}^{3-0}=1\cdot 0.1^0\cdot 0.9^3=0.729 \end{gathered}$$

$b=\left(\genfrac{}{}{0ex}{}{n}{1}\right)\cdot p^1\cdot (1-p{)}^{n-1}=(\genfrac{}{}{0ex}{}{3}{1})\cdot 0.1^1\cdot (1-0.1{)}^{3-1}=3\cdot 0.1^1\cdot 0.9^2=0.243$

$c=1-a=1-0.729=0.271$

$d=\left(\genfrac{}{}{0ex}{}{n}{n}\right)\cdot p^n\cdot (1-p{)}^{n-n}=(\genfrac{}{}{0ex}{}{3}{3})\cdot 0.1^3\cdot (1-0.1{)}^{3-3}=1\cdot 0.1^3\cdot 0.9^0=0.001$

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