Poisson Distribution- Solution Verification

Question

Suppose that the average number of lions seen on a one day safari is 5. what is the probability that tourists will see fewer than 4 lions on the next one day safari?

$\lambda= 5$

$x=0,1,2,3$

$x_1= 0$

$\frac{e^-5 \times 5^0}{0!} = .006737$

$x_2= \frac{e^-5 \times 5^1}{1!}= .033$

$x_3= \frac{e^-5 \times 5^2}{2!}= .084224$

$x_4= \frac{e^-5 \times 5^3}{3!} = .140373$

$P(x \lt 4) = P(\lambda,x_1)+ P(\lambda,x_2)+P(\lambda,x_3)+P(\lambda,x_4)=.26433$

Is this solution correct?

A life insurance salesman sells on the average 3 policies per week. What is the probability that in a given week he will sell:

a) sell some policy, (my wording, more than one): I am not sure how to tackle this question..

b) two or more but less than 5:

$\lambda=3, x=2,3,4$

$x_1= \frac{e^-3 \times 3^2}{2!}=.224041$

$x_2= \frac{e^-3 \times 3^3}{6}=.224041$

$x_3= \frac{e^-3 \times 3^4}{24}= .168031$

$P(2 \leq x \lt 5) = P(\lambda,x_1)+P(\lambda,x_2)+P(\lambda,x_3)=2(.224041)+.168031=.616113$

c) Assuming there are 5 working days in a week, what is the probability that in a given day he will sell one policy?

Using a binomial distribution $n=5 x=1 p=3/5$

$b(1,5,.60)= {5 \choose 1} \times .60 \times .40^4=.0768$

This answer seems a bit high, how should I have tackled this part? I figured since on average he sells 3 a week during a 5 day work week the average probability of success would 3/5 and since we are looking for one successful time that a policy is sold once during a 5 day week which enables the use of a binomial distribution.

Answer 1

For your first question, why did you round $$\Pr[X = 1] \approx 0.033?$$ If you do this, you will have at most $3$ decimal places of precision in your final answer, yet you report more than $3$ in your result. You should be consistent and use $0.033690$ for six places of accuracy, and your answer to six decimal places should be $0.265026$.

For your second question, compute the complementary probability for part (a): the opposite of "selling at least one policy" is "selling zero policies." So the desired probability is $$1 - \Pr[X = 0].$$ For part (b), again, you have rounding issues. The precise value rounded to six decimal places is $0.616115$ (it is actually $0.6161149710\ldots$).

For the last part (c), the idea is that if he sells $3$ policies a week and there are $5$ days in the work week, then his daily sell rate is simply a thinned Poisson process with rate $\lambda = 3/5$. Then the probability of selling exactly one policy in a given day is $$\Pr[X = 1] = e^{-3/5} \frac{(3/5)^1}{1!}.$$ The binomial distribution doesn't apply here because there aren't a fixed number of policies that are sold in the week. If you had been asked "what is the probability he sells one policy on a given day, given that he sold $n$ policies in that week," that would involve a binomial calculation.