Question

P(A or B) = 0.76
P(A or B') = 0.87
find P(A) ?
I know how to get the answer by observing only, not algebraic way. Can someone help, Please

Answer 1

\[P(A\cup B)= P(A) +P(B)-P(A\cap B) = 0.76\\(A\cup B')= P(A) +P(B')-P(A\cap B') = 0.87\]
add them together I have
\[2P(A) +1 -(P(A\cap B)+P(A\cap B')=1.63 \\2P(A) - (P(A\cap B)+P(A\cap B') =0.63\]

Answer 2

then, I have to use the Venn diagram to "see" \((P(A\cap B)+P(A\cap B') = P(A)\)
to get the answer P(A) =0.63
My question: is there any algebraic way to get it?

Answer 3

My Venn:

Answer 4

If the only problem was determining algebraically how to get the rule \(P(A)=P(A\cap B)+P(A\cap B') \) .. this actually is the Law of Total Probability (sometimes called the Partition Rule). More generally, If the events \(B_1, B_2, \ldots, B_n\) form a partition of the sample space, i.e. if \(\Large \bigcup\limits_{i=1}^n B_i=S\), where S is the sample space and \(\large B_i \cap B_j = \emptyset, \text{for }i \ne j\)(i.e. the B events are mutually exclusive), then: \[ \large P(A)=\sum_{i=1}^nP(A \cap B_i)\]
Here, \(B\) and \(B'\) are complement events, so they form a partition (P(B) + P(B')=1)

And very informal "proof" is is to use that fact that you saw this to accept via your venn diagram that \(A=(A \cap B) \cup (A \cap B')\), and now write more generally considering all the \(B_i\) events:
\( A = A \cap (B_1 \cup B_2 \cup \cdots \cup B_n)=(A \cap B_1) \cup (A \cap B_2) \cup \cdots \cup(A \cap B_n) \). I'm not too sure about a more formal proof for this.

Just a quick sketch too to support this: