40 employees in an office wear eyeglasses. 22 have single-vision correction, and 18 wear bifocals. If two employees are selected at random from this group, what is the probability that both wear bifocals? What is the probability that both have single-vision correction? Round your answers to four decimal places

Step-by-step explanation:

The number of way to select 2 employees from 40 is 40C2.

The number of way to select 2 employees from 22 is 22C2.

So the probability of the two employees having single-vision correction is just the number of ways to select the 22 employees (22C2) divided by the total number of ways to select 2 employees (40C2).

22C2 / 40C2 = 77/260

We can use the same technique for the bifocal employees.

18C2 / 40C2 = 51/260

assuming you mean addition: addition and subtraction are reverse operations.

step-by-step explanation:

73.0769230769% of the games were won