Section 5.1 Exercise 14

On the average, only 1 person in 1000 has a particular rare blood type.

Recall the Poisson distribution

\[P(X=k) = \frac{\lambda^k}{k!}e^{-\lambda}\] \[\lambda=\frac{1}{1000}=0.001\]

(a) Find the probability that, in a city of 10,000 people, no one has this blood type.

Calculate the new parameter \(\lambda'\)

\[\lambda' = 10000\lambda = \frac{10000}{1000}=10\]

\[P(X=k) = \frac{10^k}{k!}e^{-10}\]

The probability that no one has this blood type in a population of 10,000 is pretty low as can be expected.

\[P(X=0) = \frac{10^0}{0!}e^{-10} = 4.54\times10^-5 \approx 0\]

(b) How many people would have to be tested to give a probability greater than 1/2 of finding at least one person with this blood type?

Re-arranging the problem to find the complement of some of the formulation of part a, we can solve for \(\lambda\)

\[P(X \leq 1) = 1 - P(X < 1) = 1 - P(X=0) = 1 - \frac{\lambda^0}{0!}e^{-\lambda''} > \frac{1}{2}\] \[1 - \frac{1}{1}e^{-\lambda''} > \frac{1}{2}\] \[e^{-\lambda''} < \frac{1}{2}\] \[-\lambda'' < ln(\frac{1}{2})\] \[\lambda'' > ln(2) = 0.6931472\]

Recall the original rate

\[\lambda''=n \lambda = n \frac{1}{1000}\]

\[n \frac{1}{1000} > 0.6931472\] \[n > 693.1472\] So we need at least 694 people to have a 1/2 chance of finding someone with that blood type.