When do we need to use conditional probability? I am thinking about two questions.
Given that,
Student A attends class 50% of the time,
Student B attends class 60% of the time,
If student A attends class, there is 80% chance of student B attends class.
Now, given that at least one student attends class, find the probability of student A attends class.
So now, my formula is below:
Pr(A|A ∪ B)
= 0.5/((0.5+0.6 - 0.5*0.6))
=0.625

Secondly, given that at most one student attends class, find the probability of student A attends class.
And here is my calculation:
1 - Pr(B) - Pr( ∩ )
= 1 - 0.6 - 0.5*0.6
= 0.1

I feel like that the above calculation is wrong because if I add all probability, it will exceed 1 and it does not make any sense. Therefore, in this case, how can we calculate those probability? Thank you.
 A: In your calculations, you did not take the following statement into account:

If student A attends class, there is an $80\%$ chance that student B attends class.

Let $\Pr(A)$ denote the probability that student A attends class; let $\Pr(B)$ denote the probability that student B attends class.  Then we are given the following information
\begin{align*}
\Pr(A) & = 0.5\\
\Pr(B) & = 0.6\\
\Pr(B \mid A) & = 0.8
\end{align*}
The probability that both students attend class is
$$\Pr(A \cap B) = \Pr(A)\Pr(B \mid A) = (0.5)(0.8) = 0.4$$
Hence, you should have obtained
$$\Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B) = 0.5 + 0.6 - 0.4 = 0.7$$
Hence, the probability that student A attends class given that at least one student attends class is
$$\Pr(A \mid A \cup B) = \frac{\Pr(A)}{\Pr(A \cup B)} = \frac{0.5}{0.7} = \frac{5}{7}$$
As lulu indicated in the comments, you cannot assume that the probabilities of A attending class and B attending class are independent.  As you can see, the condition you did not take into account implies the event that student A attends class and student B attends class are, in fact, dependent.
Observe that the probability that only student A attends class is
$$\Pr(A) - \Pr(A \cap B) = 0.5 - 0.4 = 0.1$$
and that the probability that only student B attends class is
$$\Pr(B) - \Pr(A \cap B) = 0.6 - 0.4 = 0.2$$
The probability that student A attends class given that at most one student attends class is
$$\frac{\Pr(A) - \Pr(A \cap B)}{[\Pr(A) - \Pr(A \cap B)] + [\Pr(B) - \Pr(A \cap B)]}$$
which I will leave for you to compute.
