Question

Plant I of a manufacturing organisation employs 5 production and 3 maintenance engineers; another plant II of same organisation employs 4 production and 5 maintenance engineers. From any one of these plants, a single selection of two engineers is made. Find the probability that one them would be production engineer and the other maintenance engineer.

Answer 1

As we should choose from any of two plants, the probability that the choice will be made at the first plant is $P1=\frac{1}{2}$ , and at the second plant is $P2=\frac{1}{2}$. If we have plant I the probability to collect different two engineers is a summation of two successive events. The first event is a product of two successive events: we initially choose maintenance engineer $P1_m=\frac{3}{8}$ (at the first plant worked 8 engineers and only 3 maintenance) and then the production engineer$P1_p=\frac{5}{7}$ (at the first plant after first choice remained 7 engineers and 5 production engineers). The second event is 'we initially get production engineer' with the probability $P1^{'}_p= \frac{5}{8}$ and then some maintenance engineer with the probability $P1^{'}_m=\frac{ 3}{7}$. The overall probability for a right selection in the first plant will be $P1_{m+p}=P1_m \cdot P1_p+P1^{'}_p\cdot P1^{'}_m=\frac {3}{8}\cdot \frac {5}{7}+\frac {5}{8}\cdot \frac {3}{7}=\frac {15}{28}$ .

Similarly for second plant we get

$P2_{m+p}=P2_m\cdot P2_p+P2^{'}_p\cdot P2^{'}_m=\frac {5}{9}\cdot \frac {4}{8}+\frac {4}{9}\cdot \frac {5}{8}=\frac {5}{9}$

The probability of choosing different engineers in the organization will be

$P_{m+p}=P1\cdot P1_{m+p}+P2\cdot P2_{m+p}=\frac {1}{2}\cdot \frac {15}{28}+\frac {1}{2}\cdot \frac {5}{9}=\frac {1}{2}\cdot \frac {135+140}{252}=\frac{275}{504}$

Answer: The probability that one of two engineers will be production engineer and the other maintenance engineer is $\frac{275}{504}$ .