Probability it rains The probability it rains on Wednesday this week is 40%, while the probability it rains on Thursday this week is 30%. However, it is twice more likely to also rain on Thursday, if it has already rained on Wednesday. What is the probability it rains at least one of the two days?
Although I am not familiar with probabilities (I just did a bit of reading), I will try to start:
The probability it rains at least one of the two days is 1- the probability it will not rain on any of the two days.
This is 1-(1-40%)(1-30%)=58% 
But this is only when the two events are independent. The probability it rains on Thursday is not independent. How do I take into account this?
Thanks for your help.
 A: $p$ is the probability that it rains on Thursday if it has rained on Wednesday. $q$ is the probability that it rains on Thursday if it has not rained on Wednesday.
Given that the probability it rains on Thursday is $30\%$, we have
$$
\frac25p+\frac35q=\frac3{10}\tag1
$$
Given that it is twice as likely to rain on Thursday if it has rained on Wednesday, we have
$$
\frac{p}{1-p}=2\frac{q}{1-q}\tag2
$$
Solving $(1)$ and $(2)$ simultaneously, we get
$$
p=\frac{17-\sqrt{193}}{8},q=\frac{-11+\sqrt{193}}{12}\tag3
$$
The probability that it rains on at least one day is the complement of the probability that it doesn't rain on either day. That is,
$$
1-\frac35(1-q)=\frac{-3+\sqrt{193}}{20}\approx54.4622\%\tag4
$$
A: 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 
$P* = P(W) + P(T|{\bar W}) P({\bar W}) = P(W) + \frac12 P(T|W) P({\bar W})= 0.4 + \frac12 \frac37 0.6 =  \frac{37}{70}   \simeq 0.528$
A: P(T)=0.3=P(W)*P(T|W)+(1-P(W))*P(T|W)/2=0.4*P(T|W)+0.6*P(T|W)/2=0.7*P(T|W)
P(T|W)=0.3/0.7
P(T|$\lnot$W)=P(T|W)/2=3/14
P($\lnot$W$\land$$\lnot$T)=P($\lnot$W)(1-P(T|$\lnot$W))=0.6(1-3/14)=33/70
A: The full joint distribution of two Bernouilli random variables $W$, $T$ is given by the four probabilities $P(W, T), P(W, \overline T), P(\overline W, \overline T), P(\overline W, \overline T)$. These four numbers always satisfy the linear equation:
$$P(W, T)+P(W, \overline T)+P(\overline W, \overline T)+P(\overline W, T)=1$$
You've been given three other linear equations:
$$P(W,T)+P(W,\overline T)=0.4$$
$$P(W,T)+P(\overline W,T)=0.3$$
$$P(W,T)=0.24$$
We get the last equation from knowing that $P(T|W)=0.6$, $P(W)=0.4$ and the identity $P(A,B)=P(A|B)P(B)$.
You now have four equations for four unknowns, which can be solved for not just the answer to the question, but all possible information about the events $W$ and $T$ and the relationships between them.
A: Let's rephrase :)
The summer session has $70$ days. Over this span, $28$ exams are scheduled in the morning and $21$ in the afternoon.
However, the professors prefer to schedule double-exam days rather than spreading their activity over more "simple" days.
Let
$p$ be the number of afternoon exams scheduled in the $28$ morning exam days, and
$q$ the number of exams scheduled in the afternoons of the other $42$ days.
We know that  $p + q = 21$.
$p/28$ is the probability to have an afternoon exam after a morning exam,
$q/42$ is the probability to have an afternoon exam after a free morning.
So $p/28 = 2 \times q/42$ that gives $q = 9$
The total number of exam days is then $28 + 9 = 37$.
