The following two probabilities arise from a binomial distribution and Poisson distribution, respectively. Which gives the probability of seeing no threes's in six rolls of a standard die? $${}_6 C_2 (1/6)^2 (5/6)^4 \doteq 0.2009 \quad \quad \textrm{or} \quad \quad \frac{e^{3.5} (3.5)^2}{2!} \doteq 0.1850$$

The first one (i.e., the binomial probability), as there are a fixed number of trials and thus a limit on how many successes one can see. This is consistent with a binomial distribution, whereas in a Poisson distribution, the number of "successes" (i.e., occurrences) has no such limitation.Assume a Poisson distribution is involved and use the mean (i.e., $\lambda$) provided to find the indicated probability.

- $\lambda = 2; \ P(3) = ?$
- $\lambda = 0.3; \ P(1) = ?$
- $\lambda = 3/4; \ P(3) = ?$
- $\lambda = 1/6; \ P(0) = ?$

For each, use $\displaystyle{P(x) = \frac{e^{-\lambda} \lambda^x}{x!}}$, yielding (approximately):- $0.1804$
- $0.2222$
- $0.0332$
- $0.8465$

R: a. dpois(3,lambda=2) b. dpois(1,lambda=0.3) c. dpois(3,lambda=3/4) d. dpois(0,lambda=1/6) Excel: a. =POISSON.DIST(3,2,FALSE) b. =POISSON.DIST(1,0.3,FALSE) c. =POISSON.DIST(3,3/4,FALSE) d. =POISSON.DIST(0,1/6,FALSE)

Over the course of 365 days, 1 million radioactive atoms of Cesium-137 decayed to 977,287 radioactive atoms. Use the Poisson distribution to estimate the probability that on a given day, 50 radioactive atoms decayed.

$\lambda = (1000000 - 977287)/365 \doteq 62.227$

$\displaystyle{P(50) = \frac{e^{-62.227} 62.227^{50}}{50!} \doteq 0.0155}$R: dpois(50,lambda=62.227) Excel: =POISSON.DIST(50,62.227,FALSE)

In the last 100 years, there have been 93 earthquakes measuring 6.0 or more on the Richter scale. What is the probability of having 3 earthquakes in the same year that all measure 6.0 or more?

$\lambda = 93/100 = 0.93$. Then, $\displaystyle{P(3) = \frac{e^{-0.93} (0.93)^3}{3!} \doteq 0.0529}$R: dpois(3,lambda=0.93) Excel: =POISSON.DIST(3,0.93,FALSE)

Neuroblastoma, a rare form of malignant tumor, occurs in 11 out of a million children. In the 12,439 children of Oak Park, Illinois, 4 cases of neuroblastoma are reported. Assuming there was nothing special about Oak Park where the chance of neuroblastoma is higher than normal, find the probability of seeing more than a single case of neuroblastoma in a population this size. What can you likely conclude?

Technically this is a binomial probability problem. It is cumbersome to solve it this way unless one uses technology:

R: 1-pbinom(1,size=12439,prob=11/1000000) # results in 0.00854887 Excel: =1-BINOM.DIST(1,12439,11/1000000,TRUE)

Although, one can also get a very good approximation of this probability using the Poisson distribution (since $n$ is so large, and $p$ is so small):

First we find $\lambda = (12439)(11/1000000) = 0.136829$

Then we use the complement to find $P(X > 1) = 1 - (P(0) + P(1))$. $$1 - \frac{e^{-0.136829} (0.136829)^0}{0!} - \frac{e^{-0.136829}(0.136829)^1}{1!} \doteq 0.0085494362$$

R: 1-ppois(1,lambda=12439*11/1000000) # results in 0.008549436 Excel: =1-POISSON.DIST(1,12439*11/1000000)

Note how accurate the Poisson approximation was in this circumstance! Further, as the probabilities found are so ridiculously small, we conclude (with a significant amount of certainty) there is something special about Oak Park that makes the chance of neuroblastoma higher than 11 out of a million.

Suppose the probability of suffering a fever from the flu vaccine is $0.005$. If $1000$ people are given the vaccine, use the Poisson distribution to approximate the probability that a) 1 person suffers a fever as a result; and b) more than 6 people suffer a fever as a result.

$\lambda = (1000)(0.005) = 5$. Then, $\displaystyle{P(1) = \frac{e^{-5} 5^1}{1!} \doteq 0.0337}$

$\lambda = 5$. Now use the complement to find $P(X > 6) = 1 - P(X \le 6) = 0.2378$

R: 1-ppois(6,lambda=5) Excel: 1-POISSON.DIST(6,5,TRUE) TI-83/84: 1-poissoncdf(5,6)

Telephone calls enter a college switchboard on the average of two every three minutes. What is the probability of 5 or more calls arriving in a 9-minute period?

Poisson. $\lambda = 2 \cdot 9 / 3 = 6, \quad \quad P(X \ge 5) = 1 - P(X \le 4) \doteq 0.715$

R: 1-ppois(4,lambda=6) Excel: =1-POISSON.DIST(4,6,TRUE) TI-83/84: 1-poissoncdf(6,4)

After receiving a large shipment of computer chips, 800 chips are randomly selected. If 3 or fewer defective chips are found, the entire lot is accepted without inspecting the remaining chips in the lot. If 4 or more chips are defective, every chip in the entire lot is carefully inspected at the supplier's expense. Assume that the true proportion of nonconforming computer chips being supplied is 0.001. Use the Poisson distribution to approximate the probability the lot will be accepted.

Binomial $n=800, p=0.001$, Approximated with Poisson. $\lambda = E(X) = np = 0.8$; probability is $P(X \le 3) = P(0) + P(1) + P(2) + P(3)$, or $$e^{-0.8}\left(\frac{(0.8)^0}{0!} + \frac{(0.8)^1)}{1!} + \frac{(0.8)^2}{2!} + \frac{(0.8)^3)}{3!}\right) \doteq 0.9909$$

Exact solution (with technology):

R: pbinom(3,size=800,prob=0.001) # results in 0.9909623 Excel: =BINOM.DIST(3,800,0.001,TRUE) TI-83/84: binomcdf(800,0.001,3)

Poisson approximation (with technology):

R: ppois(3,lambda=0.8) # results in 0.9909201 Excel: =POISSON.DIST(3,0.8,TRUE) TI-83/84: poissoncdf(0.8,3)