Respuesta :
Answer:
16.15% probability that exactly 3 of them would function
Step-by-step explanation:
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Probability of each system working:
4 components, which means that [tex]n = 4[/tex]
Each has a 30% probability of failing, so [tex]p = 1 - 0.3 = 0.7[/tex]
For the system to work, all 4 components have to work. This is P(X = 4).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 4) = C_{4,4}.(0.7)^{4}.(0.3)^{0} = 0.2401[/tex]
0.2401 probability of a system working.
If you have 7 of these systems, what is the probability that exactly 3 of them would function?
Now 7 systems, so [tex]n = 7[/tex]
0.2401 probability of a system working.
We have to find P(X = 3).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 3) = C_{7,3}.(0.2401)^{3}.(0.7599)^{4} = 0.1615[/tex]
16.15% probability that exactly 3 of them would function
