Question

A complicated mechanical system contains 8 components. If there is a 18% chance that each component will fail during processing, and components fail independently of each other, then

(a) What is the probability that fewer than 3 components will fail?
(b) What is the meet likely number of failures?

Answer 1

(a) The probability that fewer than 3 components will fail is 0.8392.

(b) The mean likely number of failures is 1.44.

Explanation:

We are given that a complicated mechanical system contains 8 components. If there is an 18% chance that each component will fail during processing, and components fail independently of each other.

Let X = Number of components fail during processing

The above situation can be represented through the binomial distribution;

`P(X=r) = \binom{n}{r}* p^(r) * (1-p)^(n-r) ;x=0,1,2,.....`

where, n = number of samples (trials) taken = 8 components

r = number of success = fewer than 3 components will fail

p = probability of success which in our question is the probability

that each component will fail during processing, i.e. p = 18%

SO, X ~ Binom(n = 8, p = 0.18)

(a) The probability that fewer than 3 components will fail is given by = P(X < 3)

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

=
`\binom{8}{0}* 0.18^(0) * (1-0.18)^(8-0)+ \binom{8}{1}* 0.18^(1) * (1-0.18)^(8-1)+ \binom{8}{2}* 0.18^(2) * (1-0.18)^(8-2)`

=
`1 * 1 * 0.82^(8)+ 8 * 0.18^(1) * 0.82^(7)+28 * 0.18^(2) * 0.82^(6)`

= 0.8392

(b) The mean likely number of failures is given by the following formula;

Mean of X, E(X) = n
`*` p

=
`8 * 0.18` = 1.44

Hence, the mean likely number of failures is 1.44.