Problem 17E

A k out of n system is one in which there is a group of n components, and the system will function if at least k of the components function. Assume the components function independently of one another.

a. In a 3 out of 5 system, each component has probability 0.9 of functioning. What is the probability that the system will function?

b. In a 3 out of n system, in which each component has probability 0.9 of functioning, what is the smallest value of n needed so that the probability that the system functions is at least 0.90?

Solution 17E

Step1 of 3:

Let us consider a random variable X and it represents the number of components that function

Then X follows binomial distribution with parameters “n and p” that is X B(n, p),

The probability mass function of binomial distribution is given by

, x = 0,1,2,...,n.

Where,

n = sample size

= 5

x = random variable

p = probability of success

q = 1 - p (probability of failure)

Here our goal is:

a).In a 3 out of 5 system, each component has probability 0.9 of functioning. What is the probability that the system will function?

b).We need to find the smallest value of n needed so that the probability that the system functions is at least 0.90 when 3 out of n system, in which each component has probability 0.9 of functioning.

Step2 of 3:

a).

We have n = 5 and p = 0.9.

P(the system will function) = P(X3)

= P(X = 3)+P(X = 4)+P(X = 5)

= {+

+ }

= {10(0.7290)(0.01)+5(0.6561)(0.1)+1(0.5904)(1)}

= 0.0729 + 0.3280 + 0.5904

= 0.9913

Hence, P(the system will function) = 0.9913.

Step3 of 3:

b).

1).We have n = 3 and p = 0.9.

P(the system will function) = P(X2)

= P(X = 0)+P(X = 1)+P(X = 2)

= {+

+ }

= {1(1)(0.001)+3(0.9)(0.01)+3(0.81)(0.1)}

= 0.001 + 0.0270 + 0.2430

= 0.2710

Hence, P(the system will function) = 0.2710.

1).We have n = 4 and p = 0.9.

P(the system will function) = P(X2)

= P(X = 0)+P(X = 1)+P(X = 2)

= {+

+ }

= {1(1)(0.0001)+4(0.9)(0.001)+6(0.81)(0.01)}

= 0.0001 + 0.0036 + 0.0486

= 0.0523

Hence, P(the system will function) = 0.0523.

Hence, from (1) and (2) we have for n = 3, P(X2) = 0.2710

for n = 4, P(X2) = 0.0523

Therefore for n = 4 probability value is small.

Hence, smallest value of n is n = 4.