Question

7. (a) A factory produces steel pipes in three plant with daily production volumes of 500, 

1000 and 2000 units respectively from each of the plants. From the past experience it 

is known that the fraction of defective outputs produced by three plants are 

respectively 0.005, 0.008 and 0.010. If a pipe is selected at random from a day’s total 

production and founded to be defective, from which plant is that likely to 

have came?

Answer 1

Let event A consist in the fact that a randomly selected pipe is defective.

Let event ${B_i},\,\,i = 1,2,3$ be that the pipe was produced at the i-th plant. Then

$p({B_1}) = \frac{{500}}{{500 + 1000 + 2000}} = \frac{1}{7},\,\,p({B_2}) = \frac{{1000}}{{500 + 1000 + 2000}} = \frac{2}{7},\,\,p({B_1}) = \frac{{2000}}{{500 + 1000 + 2000}} = \frac{4}{7};$

$p(A|{B_1}) = 0.005,\,\,p(A|{B_2}) = 0.008,\,\,p(A|{B_3}) = 0.01$

Then by the formula of total probability

$p(A) = \sum {p({B_i})p(A|{B_i}) = \frac{{1 \cdot 0.005 + 2 \cdot 0.008 + 4 \cdot 0.01}}{7}} = \frac{{{\rm{0}}{\rm{.061}}}}{7} = \frac{{61}}{{7000}}$

Using the Bayes formula, we find the probabilities that the pipe came from the first, second and third plants, respectively:

$p({B_1}|A) = \frac{{p({B_1})p(A|{B_1})}}{{p(A)}} = \frac{{\frac{1}{7} \cdot 0.005}}{{\frac{{61}}{{7000}}}} = \frac{5}{{61}}$

$p({B_2}|A) = \frac{{p({B_2})p(A|{B_2})}}{{p(A)}} = \frac{{\frac{2}{7} \cdot 0.008}}{{\frac{{61}}{{7000}}}} = \frac{{16}}{{61}}$

$p({B_3}|A) = \frac{{p({B_3})p(A|{B_3})}}{{p(A)}} = \frac{{\frac{4}{7} \cdot 0.01}}{{\frac{{61}}{{7000}}}} = \frac{{40}}{{61}}$

So, it is most likely that the pipe came from the third plant.

Answer: Third plant