Calculate the probability that $x \lt 5$ given Poisson distribution states the mean is $6$ Two grocers agree that the daily demand for a particular item has Poisson distribution. However, grocer $A$ claims that the mean demand is $3$ items per day, while grocer $B$ claims that the mean demand is $6$ items per day. They agree to resolve the disagreement by observing the demand on one particular day: $B$ agrees to accept $A$’s claim if the observed demand is $4$ or less, and $A$ agrees to accept $B$’s claim if the observed demand is $5$ or more. (a) Calculate the probability that $A$’s claim is accepted when, in fact, $B$’s claim is correct

Let $X$ be the observed demand. The problem states that $X$ is Poisson distributed random variable with mean of $6$ items per day. Denote $X(\omega)$ the actual observed value of the demand, also known as realization of the random variable $X$.

The A's claim is accepted if $X(\omega) \leqslant 4$. The probability of this event is $$ \Pr\left(X \leqslant 4\right) = \sum_{n=0}^4 \Pr(X=n) $$ Can you finish this off?