Question 4.10: Worker Assignment Study A factory has three machines and eig......

(a) Given n = 8 machine operators and r = 3 machines.

Then _{8} P_{3}=\frac{8!}{(8-3!)} =\frac{8!}{5!} =\frac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{5\times 4\times 3\times 2\times 1} = 336 permutations

There are 336 distinct assignments of 8 workers to 3 machines.
(b) The probability of selecting a particular grouping of three machine operators is:

P(particular order of 3 machine operators) = \frac{1}{336} = 0.00297 (0.297%)

Any particular ordering (where order of assignment to the three machines is important) has only a very small probability (only 0.297% chance) of being selected.