I want to calculate the probability that at least in one day of the week there is more than one blackout that occurred on the same day. Secondly if it is known that at least in one day of the week a blackout occurred ,what is the probability that on Sunday occurred more than one blackout ?
For the first part i thought it was fairly simple by calculating $P\{X>0\}=1-F_X(0)$ where $F_X(0)=P\{X\leq0\}$ but then i thought it doesnt make since since that will be equaly to $1$ so my approach is wrong here.
For the second part can i go on and say that $P\{$For a blackout to occur on Sunday$\}=\frac{1}{7}$ and since the blackouts on days are independent then by the conditional probability
$P\{$A=Blackout on Sunday | B=More than one blackout occurred on same day$\}=\frac{\frac{1}{7}P(B)}{\frac{1}{7}}$
I'm assuming $\lambda=\frac{21}{2}$ is the average number of blackouts in a week.
Then, the average number of blackouts in one day is just $\frac{\lambda}{7}=\frac{3}{2}$, since it is a Poisson process, and thus the average number of events is proportional to the observation interval.
So if you let $X$ be the number of blackouts in a day, then $$P(X=k)= e^{-3/2}\frac{\left(\frac{3}{2}\right)^k}{k!}, \ \ k=0,1,2,\dots,$$ and the probability of having more than one blackout in one day is $$p=1-P(X\leq 1)= 1- P(X=0)-P(X=1)=1-\frac{5}{2}e^{-3/2}.$$
Now, getting to the week, let $Y$ be the number of days in which you have more than one blackout. $Y$ has a Binomial distribution with parameters $n=7$ and $p$ given by the previous equation. So
$$P(Y=k)={7\choose k} p^k(1-p)^{7-k}, \ \ 0\leq k \leq 7,$$ and the answer to your question is $$P(Y\geq 1) = 1-P(Y=0)=1-(1-p)^7=1-\left(\frac{5}{2}\right)^7e^{-21/2}.$$