Question

A Supervisor of a pharmacy found that 30% of the workers receive their spending money from part – time jobs. If five workers are selected at random, find the probability that at least 3 of them will have part – time jobs.

Answer 1

Solution.

At least 3 of workers means 3 over 5 workers or 4 over 5 workers or 5 over 5 workers. These events are independent, let's calculate probability of them independently and then just add all probabilities.

Consider when 3 over 5 workers have part-time jobs. Probability of it is

$p_3 = (30\%)^3\cdot(70\%)^2 \cdot C_5^3 =0.027\cdot0.49\cdot10 = 0.1323$

Here $(30\%)^3\cdot(70\%)^2$ is a probability of 1 permutation: part-time worker, part-time worker, part-time worker, full-time worker, full-time worker. There are $C_5^3$ permutations.

Consider when 4 over 5 workers have part-time jobs. Probability of it is

$p_4 = (30\%)^4\cdot(70\%)^1 \cdot C_5^4 = 0.0081\cdot0.7\cdot5 = 0.02835$

Consider when 5 over 5 workers have part-time jobs. Probability of it is

$p_5 = (30\%)^5 \cdot C_5^5 = 0.00243 \cdot 1 = 0.00243$

Sum of probabilities is

$p = p_3+p_4+p_5 = 0.1323 + 0.02835 + 0.00243 = 0.16308$

Answer: 0.16308