Let $X$ represent the number of emails you receive in an hour on a weekday and let $Y$ represent the number of emails you receive in an hour on a weekend. Let $Z$ be the random variable representing how many emails you received in an hour knowing that the day you selected could be any day of the week.
$$X\sim\mathrm{Po}(10)\implies\operatorname{P}(X=x)=\frac{10^x}{e^{10}x!}\tag{1}$$
$$Y\sim\mathrm{Po}(2)\implies\operatorname{P}(Y=y)=\frac{2^y}{e^{2}y!}\tag{2}$$
Bayes’ theorem tells us
$$\operatorname{P}(B \,|\, A) = \frac{ \operatorname{P}(B) \, \operatorname{P}(A \,|\, B) }{\operatorname{P}(B) \, \operatorname{P}(A \,|\, B) + \operatorname{P}\left(B'\right) \, \operatorname{P}\left(A \,\middle|\, B'\right)}$$
Taking $B$ to be the event $D$ and $A$ to be the event $Z=0$, we have
$$\operatorname{P}(D \,|\, Z=0) = \frac{ \operatorname{P}(D) \, \operatorname{P}(Z=0 \,|\, D) }{\operatorname{P}(D) \, \operatorname{P}(Z=0 \,|\, D) + \operatorname{P}\left(D'\right) \, \operatorname{P}\left(Z=0 \,\middle|\, D'\right)}$$
Given if $D$ occurs, $Z=0$ can actually be more precisely written as $X=0$; if $D'$ occurs, then $Z=0$ can actually be more precisely written as $Y=0$, giving you
$$\operatorname{P}(D \,|\, Z=0) = \frac{ \operatorname{P}(D) \, \operatorname{P}(X=0 \,|\, D) }{\operatorname{P}(D) \, \operatorname{P}(X=0 \,|\, D) + \operatorname{P}\left(D'\right) \, \operatorname{P}\left(Y=0 \,\middle|\, D'\right)}$$
By the definition of our variables, we can further contract this to
$$\operatorname{P}(D \,|\, Z=0) = \frac{ \operatorname{P}(D) \, \operatorname{P}(X=0) }{\operatorname{P}(D) \, \operatorname{P}(X=0) + \operatorname{P}\left(D'\right) \, \operatorname{P}(Y=0)}$$
because if we’re analyzing $X$, we know we’re looking at a weekday, and if we’re analyzing $Y$, we know we’re looking at a weekend.
Use $(1)$ and $(2)$ in conjunction with $\operatorname{P}(D)=5/7$ and $\operatorname{P}\left(D'\right)=2/7$ to arrive at your answer.
The point in Probabilities is the following: Denote $W$ the event that it rains on Wednesday, ${\bar W}$ the event that it doesn't. Likewise with $T$ and $\bar T$ for Thursday.
Then $P(W) =0.4$.
By marginalization, you have $0.3 = P(T) = P(T|W) P(W) + P(T|{\bar W}) P({\bar W}) $. So if you are interested in conditional events like $P(T|W)$, you could use this formula as follows. The additional info is that $ P(T|W) = 2 P(T|{\bar W}) $. So you obtain $0.3 = P(T|W) 0.4 + \frac12 P(T|W) 0.6$. This allows you to calculate $P(T|W) = 3/7$.
Now the situation $P*$ that it rains on any of the two days can be split into two disjoint events, so
$P* = P(W) + P(T|{\bar W}) P({\bar W}) = P(W) + \frac12 P(T|W) P({\bar W})= 0.4 + \frac12 \frac37 0.6 = \frac{37}{70} \simeq 0.528$
Best Answer
The set-up is equivalent to
$X$ number of heads are obtained when a coin is tossed $n$ times , when the probability of obtaining head is "p". I hope you know that, X follows a binomial distribution with parameters "n" and "p".
As according to your question, we have, $n = 5,x=1,p=\frac{1}{2}=(1-p)$
$$Pr[X \geq x ] = Pr[X \geq 1 ] = 1 - Pr[X=0] = 1 - \frac{1}{2^5} $$