Question

At a software company, skilled workers have been hired for a project. Out of 75 candidates, 48 of them were software engineer; 35 of them were hardware engineer; 42 of them were network engineer; 18 of them had skills in all three jobs and all of them had skills in at least one of these jobs. How many candidates were hired who were skilled in exactly 2 jobs?

(a) 69

(b) 14

(c) 32

(d) 8

This question was addressed to me during an online interview.

This is a very interesting question from Discrete Probability in chapter Discrete Probability of Discrete Mathematics

Answer 1

Right option is (b) 14

The best explanation: Since 18 are skilled in all 3. Subtract 18 from all three to get a total with single skilled and double skilled workers including the duplicates. Software engineers = 48 – 18 = 30, Hardware engineers = 35 – 18 = 17, Network engineers = 42 – 18 = 24 making a total of 71 and this is a total set of single and double skilled workers including duplicates. Out of 75 candidates, 18 were skilled in three areas. So, 75 – 18 = 57 (actual no of workers skilled with single and both skills) Now the difference between the number without duplicates (57) and with duplicates (71), 71 – 57 = 14. So, 14 are skilled in exactly two jobs.