Math Problem, please help.

A company finds that one out of five employees will be late to work on a given day. If this company has 41 employees, find the probabilities that the following number of people will get to work on time.

a) Exactly 27 workers or exactly 31 workers.

The answer is .1971, but I just don't know how to get it. I tried binompdf for 27 + binompdf for 31 but I'm not getting it.

1/5 are late
.2 are late
.8 are not late

P(27 are on time) = 41C27 * .8^27 * .2^14 = 0.01396006273363827217623063047936
P(31 are on time) = 41C31 * .8^31 * .2^10 = 0.1136929869973370117524918621999

P(27 or 31) = 0.12765304973097528392872249267926
12.77%

one in five employees are late
1/5 or 20% of the employees are late
80% of the employees are on time

there are 41 employees
27 are on time
there are 41C27 ways that 27 employees are on time
41C27 = 35240152720
35240152720 * .8^27 * .2^14 = 0.01396006273363827217623063047936
or about a 1.4% probability that 27 are on time and 14 are late

41C31 = 1121099408
1121099408 * .8^31 * .2^10 = 0.1136929869973370117524918621999
or about a 11.4% probability that 31 are on time and 10 are late

so the probability of
27 or 31 = 1.4 + 11.4
P(27 or 31 are on time) = 12.8%