Question

A rail company knows that on average they will have 3 people ill each day. In order to make plans for replacements they have asked you to calculate the following probabilities: In a single day - (1) No employees ill (2)Two or less employees ill (3) Four or more employees ill

Answer 1

The unknown probabilities depend on distribution of number of number of ill patients.
But in such situations the most frequently is used Poisson distribution.
1) No employees ill, X = 0, lambda=3
P(0) = e^-3 3⁰/0! = 0.5*1/1 = .05

2)Two or less ill
P(X<=2) = P(0) + P(1) + P(2) = .15

P(1) = e^-3 3¹/1! = 0.05*3/1 = 0.15
P(2) = e^-3 3²/2! = 0.05*9/2 = 0.22
P(X<=2) = P(0) + P(1) + P(2)= .05 +.15 + .22 =.42

3)Probability of four or more employees ill

P(X>=4) = 1 – P(X<=3) = 1 – (P(0) + P(1) + P(2) + P(3))
P(3) = e^-3 3³/3! = 0.05*27/6 = 0.27
P(X>=4) = 1 – P(X<=3) = 1 – (P(0) + P(1) + P(2) + P(3)) = 1 – (.05 +.15 +.22 +.27) = 1 - .69 =.31