A space shuttle requires 3-out-of-4 of its main engines to achieve orbit. Each engine has a reliability of 0.97. What is the probability of achieving orbit?

Nov 22, 2016

$.99$

Explanation:

if each engine has a reliability of .97 then we can look at this $p \left(x\right) = .97$. This is per engine.

Now if we only need 3 out of 4 then this converts into a Binomial distribution which tells us the probability of $k$ successes out of n trials.

f(x,n,k) =(""_k^n) p(x)^k(1-p(x))^(n-k)
in this case we know that we need at least 3 so

sum_(i=3)^4f(x,4,i) =sum_(i=3)^4(""_i^4) .97^i(.03)^(4-i) = .99