Question

In the summer months, an ecotourism company has 23 employees (10 permanent and 13 seasonal). Four (4) employees were sick with the flu on July 15th, 2003.
What is the probability that at least 3 of the 4 sick employees were seasonal?

Answer 1

Number of ways to select r items from n, nCr = n!/(r! x (n-r)!)

Total number of employees = 23

Number of permanent employees = 10

Number of seasonal employees = 13

Number of ways in which any 4 people can be selected from 23 = 23C4 = 8,855

P(at least 3 of the 4 sick employees were seasonal) = P(3 of the 4 sick employees were seasonal) + P(all of the 4 sick employees were seasonal)

= (Number of ways to select 3 seasonal employees from 13 and 1 permanent employee from 10 / Total number of selections possible) + (Number of ways to select 4 seasonal employees from 13 / Total number of selections possible)

= (13C3 x 10C1 / 23C4) + (13C4/23C4)

= (286 x 10 / 8,855) + (715 / 8,855)

= 0.4037