Chapter 2 Section 3 Question 19

Question

Suppose an urn contains 1000 balls - one of these is black, and the other 999 are white. Suppose that 100 balls are randomly drawn from the urn with replacement. Use the appropriate Poisson distribution to approximate the probability that five black balls are observed.

Answer

Author: Mohammad-Ali Bandzar| Date:Oct 19 2020

From Example 2.3.6 in the book we know: $$p\scriptsize{Y}\normalsize{(y)=P(Y=y)=\frac{\lambda^y}{y!}e^{-\lambda}}$$ $\lambda$ represents the mean number of occurrences in the given interval in this case it will be the number of draws multiplied the probability of drawing a black ball $\lambda=100*\frac{1}{1000}=0.1$

$y$ in this case will be the desired number of black balls drawn, in this case $y=5$ $$P(Y=y)=\frac{\lambda^y}{y!}e^{-\lambda}$$ $$P(Y=5)=\frac{0.1^5}{5!}e^{-0.1}$$ $$P(Y=5)\approx 0.0000000754031$$

From Example 2.3.6 in the book we know: $$p\scriptsize{Y}\normalsize{(y)=P(Y=y)=\frac{\lambda^y}{y!}e^{-\lambda}}$$ $\lambda$ represents the mean number of occurrences in the given interval in this case it will be the number of draws multiplied the probability of drawing a black ball $\lambda=100*\frac{1}{1000}=0.1$

$y$ in this case will be the desired number of black balls drawn, in this case $y=5$ $$P(Y=y)=\frac{\lambda^y}{y!}e^{-\lambda}$$ $$P(Y=5)=\frac{0.1^5}{5!}e^{-0.1}$$ $$P(Y=5)\approx 0.0000000754031$$