Answer :
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;
[tex]P(X=r) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ;x=0,1,2,.....[/tex]
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)
= [tex]\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}[/tex]
= [tex]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}[/tex]
= 0.8392
(b) The mean likely number of failures is given by the following formula;
Mean of X, E(X) = n [tex]\times[/tex] p
= [tex]8 \times 0.18[/tex] = 1.44
Hence, the mean likely number of failures is 1.44.