Answer:
A) 0.335; B) 0.343; C) 0.599
Step-by-step explanation:
For part A,
To find the probability that the user has more than one infection per year, we can either add together the probabilities for 2, 3, 4 and 5; or we can add together the probabilities for 0 and 1 and subtract them from 1:
1-(P(X = 0)+P(X = 1))
= 1-(0.343+0.322) = 1-0.665 = 0.335
For part B,
The probability that the user has no infections per year is P(X = 0); this is 0.343.
For part C,
The probability that the user has between 1 and 3 infections (inclusive) per year is
P(1 ≤ X ≤ 3) = 0.322+0.201+0.076 = 0.599