Answer
$\color{blue}{e^{-0.7} \approx 0.50$}$
Work Step by Step
Let $Y$ be the event that it will be at least one week or seven (7) days from now until the next death occurs.
This event is the same as the event in which there are no deaths in the next seven (7) days.
Now, the probability of no death in any one day is given by $P(X=0) = \dfrac{e^{-0.1}
(0.1)^0}{0!} = e^{-0.1}$, where $X$ denotes the number of deaths in a day and has a Poisson distribution with rate parameter $\lambda = 0.1$ deaths per day, i.e., the pdf of $X$ is $p_X(k) = \dfrac{e^{-0.1}(0.1)^k}{k!},\ k=0,1,2,3,\ldots.$
Thus, the probability of no deaths in the next seven (7) days is thus $(P(X=0))^7 = (e^{-0.1})^7 = e^{-0.7} \approx 0.49658530 \approx \color{blue}{0.50.}$
