Answer:
(a) The probability that the system will function is 0.9914.
(b) The expected number of components that function is 4.5.
(c) The smallest possible value of <em>n</em> is 4 for which the probability of the system function is at least 0.90.
Step-by-step explanation:
It is provided that a <em>k</em> out of <em>n</em> system will function if at least <em>k</em> components will function out of <em>n</em>.
The distribution of the number of components functioning is Binomial.
Then the probability mass function of the number of components working in a 3 out of 5 system is:
(a)
The probability of a component functioning is, P (X) = <em>p</em> = 0.90.
Now for the system to function at least 3 components must function.
Compute the probability of P (X ≥ 3) as follows:
P (X ≥ 3) = 1 - P (X < 3)
= 1 - P (X = 0) - P (X = 1) - P (X = 2)
Thus, the probability that the system will function is 0.9914.
(b)
The expected value of a Binomial random variable is:
Compute the expected number of components that function as follows:
Thus, the expected number of components that function is 4.5.
(c)
- In part (a) we computed the probability of a 3 out of 5 system working as 0.9914.
- Consider a 3 out of 4 system.
Check whether the probability of a 3 out of 4 system functioning is at least 0.90.
Compute the probability of a 3 out of 4 system working as follows:
P (X ≥ 3) = 1 - P (X < 3)
= 1 - P (X = 0) - P (X = 1) - P (X = 2)
The probability of a 3 out of 4 system working is 0.9477.
- Consider a 3 out of 3 system.
Check whether the probability of a 3 out of 3 system functioning is at least 0.90.
Compute the probability of a 3 out of 3 system working as follows:
P (X ≥ 3) = 1 - P (X < 3)
= 1 - P (X = 0) - P (X = 1) - P (X = 2)
The probability of a 3 out of 3 system working is 0.7290.
Thus, the smallest possible value of <em>n</em> is 4 for which the probability of the system function is at least 0.90.