probability of A or B or A and B
I am reading Probability tutorial and found this question and answer.
A student goes to the library. The probability that she checks out (a) a work of fiction is 0.40, (b) a work of non-fiction is 0.30, and (c) both fiction and non-fiction is 0.20. What is the probability that the student checks out a work of fiction, non-fiction, or both?
Solution: Let F = the event that the student checks out fiction; and let N = the event that the student checks out non-fiction. Then, based on the rule of addition:
P(F ∪ N) = P(F) + P(N) - P(F ∩ N) P(F ∪ N) = 0.40 + 0.30 - 0.20 = 0.50
According to me, as it is told to "checks out a work of fiction, non-fiction, or both" so probability is "Fiction union Non-fiction union both" which it should be ( F u N u (F n N) ) but it is told as (fiction union non-fiction) is answer.
I got a reference from this link here in answer section
P(A or B)=P(A)+P(B)-P(A & B)=0.5
P(A or B or (A & B))
=P((A or B) or (A & B))
=P(A or B)+P(A & B)-P((A or B)&(A & B))
=P(A or B)+P(A & B)-P(A & B) --> how P((A or B)&(A & B)) became P(A & B) here
=P(A or B)=0.5 4 Answers
$\begingroup$According to me, as it is told to "checks out a work of fiction, non-fiction, or both" so probability is "Fiction union Non-fiction union both" which it should be ( F u N u (F n N) ) but it is told as (fiction union non-fiction) is answer.
The symbol $\cup$ stands for "union" (a.k.a. "disjunction") of the events, and it means "inclusive or". It includes the intersection (a.k.a "conjunction", symbol $\cap$). So "fiction union non-fiction" is "fiction, or non-fiction, or both".
The symbol $\oplus$ stands for "exclusive disjunction", or "xor". It excludes the intersection. So "fiction xor non-fiction" would be "either fiction or non-fiction but not both"
tl;dr $(F\cup N)\cup(F\cap N)\\ = (F\cup N\cup F)\cap(F\cup N\cup N)\\ = (F\cup N)\cap(F\cup N)\\ = F\cup N$
$\endgroup$ $\begingroup$Assuming the choice is random and the events are independent, and using your notation:
$$\begin{cases}P(F)=0.4\\{}\\P(N)=0.3\\{}\\P(F\cap N)=0.2\end{cases}\;\;\implies\;\;P(F\cup N)=P(F)+P(N)-P(F\cap N)=0.5$$
Observe that $\;F\cup N\;$ is the event that either fiction or non fiction or both happens (i.e., is chosen). Unless otherwise explicitly stated, "or" in mathematics is always inclusive.
$\endgroup$ 4 $\begingroup$Do you mean that P((F ∪ N) ∪ (F ∩ N)) should be the LHS of the equation?
But if some event x is in (F ∩ N), then it is in F and in N and therefore it is also in (F ∪ N). So why do you want to take the union of the two sets?
$\endgroup$ 1 $\begingroup$If we formulate probabilities over sets of outcomes, then you can use the algebraic laws of $\cap$ and $\cup$:
$$ P((F \cup N) \cup (F \cap N)) $$
Associate on $\cup$: $(A \cup B) \cup C = A \cup (B \cup C)$
$$ P(F \cup (N \cup (F \cap N)) $$
Commute on $\cap$: $A \cap B = B \cap A$
$$ P(F \cup (N \cup (N \cap F)) $$
Absorbtion on $\cup$ over $\cap$: $A \cup (A \cap B) = A$
$$ P(F \cup N) $$
$\endgroup$
