Question

a). In a certain population an average of 6 new cases of esophageal cancer are diagnosed each year. If the annual incidence of esophageal cancer follows a Poisson distribution, find the probability that in a given year the number of newly diagnosed cases of esophageal cancer will be at least 3.

b). In a family of 4 children, what is the probability that at least 3 of the children are boys?

Answer 1

a)

poisson probability distribution

P(X=x) = e-λλx/x!

λ =    6

P(at least 3) = 1 - P(X=0) - P(X=1) - P(X=2)

P ( X =    0   ) = e^-6*6^0/0!=   0.00248
P ( X =    1   ) = e^-6*6^1/1!=   0.01487
P ( X =    2   ) = e^-6*6^2/2!=   0.04462

P(at least 3) = 1 - P(X=0) - P(X=1) - P(X=2) = 1 - 0.0620 = 0.9380

b)

n=4

P(boy) = P(girl) = p = 0.50

P ( X =    3   ) = C(4,3) * 0.5^3 * (1-0.5)^1 =            0.2500
P ( X =    4   ) = C(4,4) * 0.5^4 * (1-0.5)^0 =            0.0625

P(at least 3) = P(X=3) + P(X=4) = 0.25 + 0.0625 = 0.3125