Question

. The probability of contracting pulmonary tuberculosis in a given area is 0.003. What

is the likelihood that an examination of 1000 people will reveal three patients?

Answer 1

When the value $n$ in a Binomial distribution is large and the value of $p$ is very small, the Binomial distribution can be approximated by the Poisson distribution

If,

$n\gt20\\ np\lt5 \space or\space n(1-p)\lt5$

then the Poisson distribution is a good approximation.

For this case,

$n=1000\gt20\\ p=0.003\\ np=(1000\times0.003)=3\lt5$

Therefore, the conditions above hold and the Poisson distribution is a good approximation.

The Poisson distribution to be used has parameter $\lambda=np=3$.

We determine the probability $P(X=3)={e^{-3}3^3\over 3!}={e^{-3}\times27\over 3!}= 0.2240418$

Thus, the likelihood that an examination of 1000 people will reveal three patients is  0.2240418.