Question

1. My family really likes oatmeal. On average, we eat oatmeal 4 times each week. The probability of us eating oatmeal is the same for any fixed length of time, and eating oatmeal at any point does not impact the odds of us eating oatmeal at another point. (Our oatmeal eatings are independent.)
   1. What is the probability that we eat oatmeal 4 times in one week?
   2. What is the probability that we eat oatmeal twice or fewer in one week?
   3. What is the probability that we eat oatmeal once on any given day?
   4. Assume that we could only eat oatmeal once on a given day or not at all. Given this restriction, what is the probability that we would eat oatmeal?
   5. How do answers (c) and (d) compare with one another?

Answer 1

Let X denotes the number of times oatmeal has eaten each week.

X ~ Poisson(4)

The probability mass function of X is

$P(X=x) = \frac{e^{-4}*4^x}{x!},x=0,1,2,3,...$

a) The probability that we eat oatmeal 4 times in one week

$=P(X=4) = \frac{e^{-4}*4^4}{4!} = 0.195367$

b) The probability that we eat oatmeal twice or fewer in one week

$=P(X\leq 2) =\sum_{x=0}^{2} \frac{e^{-4}*4^x}{x!} = 0.238103$

c) Let Y denotes the number of times oatmeal has eaten on any given day.

Y ~ Poisson(4/7) or Y ~ Poisson(0.5714)

The probability mass function of Y is

$P(Y = y)=\frac{e^{-0.5714}*0.5714^y}{y!}, y = 0,1,2,3,...$

The probability that we eat oatmeal once on any given day

$=P(Y = 1)=\frac{e^{-0.5714}*0.5714^1}{1!} = 0.322689$