# 2 component system vs 4 component system, number of combos to consider

My book has the question

"One design for a system requires the installation of two identical components. The system will work if at least one of the components works. An alternative design requires four of these components, and the system will work if at least two of the four components work. If the probability that a component works is 0.9, and if the components function independently, which design has the greater probability of functioning?"

I know p*p for and and p+p for or, so for the two component system with only one needing to function you have together 0.9 chance for one plus 0.9 probability for the other total chance for it to work, so 1.8.

For the 4 component system I initially started to do 0.9*0.9+0.9*0.9, but then realized that would only be/cover two combos, such as (if have components a b c d) a&b or c&d, not something like a&c b&d.

So would I have to do that for every combo like:

a 0.9 * b 0.9 + c 0.9 * d 0.9 + a 0.9 * c 0.9 + b 0.9 * d 0.9 + a 0.9 * d 0.9 + c 0.9 * b 0.9

Or is there some simpler trick to this I"m missing?

Edit: Will wait.. you shouldn't have p above 1 so... all that's off?

• No probability can exceed $1$ so your $1.8$ must be wrong. You can usually only simply add probabilities for mutually exclusive events (while independence allow you to simply multiply them) Sep 12 '19 at 0:02

The total chance for something to work can never be more than $$1$$. If you have two items with chance of $$0.9$$ each, the only way it can fail is if both components fail. Assuming independence, the chance of that is $$0.1^2=0.01$$, so the chance of success is $$0.99$$

If you want to list all the possibilities for two of four working, there are $$2^4=16$$ possibilities, of which $$11$$ have at least two working. Because of the symmetry, you can group all the possibilities with a given number working, so there are $${4 \choose 2}=6$$ possibilities with exactly two working. What is the chance of that?

• Nice method (+1). Especially if binomial distribution not yet covered. Sep 12 '19 at 0:12

First, you first answer can't be correct because probabilities can't be greater than 1 (and you have a bogus probability of 1.8').

The 2-component system fails only if both fail, so it works with probability $$1 - (.1)^2 = .99.$$

The 4-component system works if 2 or more components work. Let the number of working components be $$X \sim \mathsf{Binom}(4, .9).$$ Then you can sum terms of the this binomial PDF to find $$P(X \ge 2) = 0.9963.$$ You will have fewer terms to sum if you find $$P(X \ge 2) = 1 - P(X \le 1).$$ [Computations in R below.]

sum(dbinom(2:4, 4, .9))
[1] 0.9963
1 - pbinom(1, 4, .9)
[1] 0.9963
1 - .1^4 - 4*.9*.1^3
[1] 0.9963
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