According to a 2007 CBC article, Canadians face about a 10% chance of acquiring an infection when hospitalized. If records of 15 randomly selected hospitalized patients are examined find for the number of Canadians who are infected when hospitalized. d. the probability that two or more of the 15 aquire an infection while hospitalized Solution Given X~Binomial(n=15, p=0.1) P(X=x)=xC15*(0.1^x)*(0.9^(15-x)) So the probability is P(X>=2)=1-P(X=0)-P(X=1)= 0.450957.

1. According to a 2007 CBC article, Canadians face about a 10% chance of acquiring an infection
when hospitalized. If records of 15 randomly selected hospitalized patients are examined find for
the number of Canadians who are infected when hospitalized.
d. the probability that two or more of the 15 aquire an infection while hospitalized
Solution
Given X~Binomial(n=15, p=0.1) P(X=x)=xC15*(0.1^x)*(0.9^(15-x)) So the
probability is P(X>=2)=1-P(X=0)-P(X=1)= 0.450957