Question

an inspection procedure at a manufacturing plant involves picking three items at random and then accepting the whole lot if at least two of the three items are in perfect condition. if in reality 80% of the whole lot is perfect, what is the probability that the lot will be accepted?

(a) .512
(b) .560
(c) .640
(d) .896
(e) .992

Answer 1

Answer: (d) .896

Step-by-step explanation:

In binomial distribution, the probability of getting success in x trials is given by :-

$P(X=x)=^nC_xp^{x}(q)^{n-x}$

, where  n is the total number of trials , p is the probability of getting success in each trial and q is the probability of not getting success in each trial  .

Given : The proportion of the whole lot is perfect =0.80

Let x be the number of items found in perfect condition.

i.e. p=0.80 and q= 1-0.8=0.20

n= 3

The whole lot will be accepted if at least two of the three items are in perfect condition.

Then, the probability that at least two of the three items are in perfect condition will be :-

$P(X\geq2)=P(2)+P(3)\\\\=^3C_2(0.8)^{2}(0.2)^{1}+^3C_3(0.8)^{3}(0.2)^{0}$

$=(3)(0.64)(0.2)+(1)(0.512)\\\\=0.896$

Hence, the  probability that the lot will be accepted is 0.896.

Thus , the correct answer is (d) .896.