The question goes as such: An event A can occur if only one of the mutually exclusive events B1, B2, or B3 occur. Show that
P(A) = P(B1)P(A|B1)+P(B2)(A|B2)+P(B3)*(A|B3)
my working out:
P[A|(B1 U B2 U B3)] = P[A INTERSECT (B1 U B2 U B3)] / P(B1 U B2 U B3)
Confused how to move on...
can someone please explain how Bayes Theorem works, and then how I should follow through this question.
On
Mutually exclusive events means that $P(B_i \cap B_j) = 0$ for any $i \ne j$, as such the partition of the event space (lets say $\Omega_A = B_1 \cup B_2 \cup B_3$) is complete and any event $A$ (on this sub-event space) can be written as:
$A = A \cap \Omega_A = A \cap ( B_1 \cup B_2 \cup B_3)$ then by total probability formula one has:
$$P(A) = P(A \cap \Omega_A) = P(A \cap ( B_1 \cup B_2 \cup B_3))$$
since $B_i$'s are mutually exclusive
$$P(A) = P(A \cap B_1) + P(A \cap B_2) + P(A \cap B_3)$$
which by the conditnal probability formula gives:
$$P(A) = P(B_1)P(A | B_1) + P(B_2)P(A | B_2) + P(B_3)P(A | B_3)$$
it all depens on the fact that the $B_i$'s constitute a non-overlapping (mutually exclusive) partition of the (sub-)event space $\Omega_A$, as such requiring that the event $A$ occurs if only one of the mutually exclusive events $B_i$ occurs is equivalent to the joint event $A \cap \Omega_A$
HINT
Since the event $A$ requires any one of the events $B1, B2, B3$ to occur, we have :
$P\left((B1\cup B2\cup B3) ~|~ A\right) = 1$
$\dfrac{P((B1\cup B2\cup B3) ~\cap~ A)}{P(A)} = 1 \implies P(A) = P((B1\cup B2\cup B3) ~\cap~ A) $