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

Answer:

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

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

Step-by-step 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;

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

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)

= $\binom{8}{0}\times 0.18^{0} \times (1-0.18)^{8-0}+ \binom{8}{1}\times 0.18^{1} \times (1-0.18)^{8-1}+ \binom{8}{2}\times 0.18^{2} \times (1-0.18)^{8-2}$

= $1 \times 1 \times 0.82^{8}+ 8 \times 0.18^{1} \times 0.82^{7}+28 \times 0.18^{2} \times 0.82^{6}$

= 0.8392

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

      Mean of X, E(X) = n $\times$ p

                                 = $8 \times 0.18$ = 1.44

Hence, the mean likely number of failures is 1.44.