multiple probability - statistics I have 3 machines independently running electrical resistors and I run at once.
The probability of a 5-ohm resistor is 0.2 and the probability of 10 ohm is 0.5 and 15 ohm is 0.2
What is the probability after one run of each at least one resistor is 15 ohm? 
I know that P( at least 1 15 ohm resistor) = 1- P(all not 15 ohm) 
How do I find the P(all not 15 ohm)? 
I know the probability of all 15 ohm is 0.008 (0.2)^3.  Could I take the complement of 1-0.008? 
 A: The number of 15ohm resistors in 3 is
$X \sim \mathsf{Binom}(n=3, p = 0.2).$
(a) You want $P(X \ge 2) = P(X=2)+P(X=3) =  0.104.$
Computation using R statistical software where dbinom is a binomial PDF.
sum(dbinom(2:3, 3, .2))
[1] 0.104

(b) $P(X \le 2) = P(X=0) + \cdots P(X=2) = 0.992.$ In R, pbinom is a binomial CDF.
pbinom(2, 3, .2)
[1] 0.992
1 - dbinom(3, 3, .2)
[1] 0.992

There may be some confusion between 'not all' above and 'all not' (not exactly standard English), which might mean getting
no 15ohm resisters at all. Then
$P(X = 0) = (1-.2)^3$ $= (0.8)^3$ $= 0.512.$

Here is a plot of the PDF of $\mathsf{Binom}(3, .2).$
x = 0:3;  pdf = dbinom(x, 3, .2)
plot(x, pdf, type="h", lwd=3, col="blue", 
     main="PDF of BINOM(3, .2)")
 abline(h=0, col="green2")
 abline(v=0, col="green2")


I have (mostly) used R statistical software to get the exact
probabilities. Maybe you are expected to use the binomial
PDF formula. (Sometimes it is possible to use a normal
approximation to get nearly-correct binomial probabilities,
but I would not advise using a normal approximation here.)
