Respuesta :
Answer:
Probability that no more than 2 out of 12 parts tested are defective is 0.8891.
Step-by-step explanation:
We are given that the probability that a machine part is defective is 0.1.
Twelve parts are selected at random.
The above situation can be represented through binomial distribution;
[tex]P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,.......[/tex]
where, n = number trials (samples) taken = 12 parts
r = number of success = no more than 2
p = probability of success which in our question is probability that
a machine part is defective, i.e; p = 0.1
Let X = Number of machine parts that are defective
So, X ~ Binom(n = 12, p = 0.1)
Now, Probability that no more than 2 out of 12 parts tested are defective is given by = P(X [tex]\leq[/tex] 2)
P(X [tex]\leq[/tex] 2) = P(X = 0) + P(X = 1) + P(X = 2)
= [tex]\binom{12}{0} \times 0.1^{0} \times (1-0.1)^{12-0}+\binom{12}{1} \times 0.1^{1} \times (1-0.1)^{12-1}+\binom{12}{2} \times 0.1^{2} \times (1-0.1)^{12-2}[/tex]
= [tex]1\times 1 \times 0.9^{12}+ 12 \times 0.1^{1} \times 0.9^{11}+ 66\times 0.1^{2} \times 0.9^{10}[/tex]
= 0.8891
Therefore, probability that no more than 2 out of 12 parts tested are defective is 0.8891.
