A and B are independent events such that P(A)=0.7, P(B)=k, P(A U B)=0.8 Find the value of k.
Solution given:
P(A U B) = P(A) + P(B) - P(A ∩ B).....(i) [Addition Rule]
P(A ∩ B) = P(A).P(B).....(ii) [Multiplication rule]
Substituing (ii) in (i),
P(A U B) = P(A) + P(B) - P(A).P(B)
0.8 = 0.7 + k - 0.7k
0.1 = 0.3k
Hence k = 1/3
I find this solution wrong because I think we cannot just substitue the multiplication rule in the addition rule because the intersections in both the rules mean different things.
In the additon rule, intersection refers to an outcome that satisfies both the events.
In multiplication rule, intersection refers to one event happening after another.
Example: If A={2,3} and B={3,9}
Addition rule: A∩B = {3}
Multiplication rule: A∩B = {(2,3), (2,9), (3,3), (3,9)}
Intersection means the same thing in both cases: That the intersection $A\cap B$ occurs means just that both events occur, or if you like, the outcome "satisfies both".
If $A=\{2,3\}$ and $B=\{3,9\}$ then $A\cap B = \{3\}$ and $A\times B = \{(2,3),(2,9),(3,3),(3,9)\}$. This latter product is the Cartesian product, not the intersection.
Suppose an experiment yields $1$, $2$, or $3$, and an independent replication also yields $1$, $2$, or $3$. Then the set of possible outcomes of the sequence of two experiments is the Cartesian product: $$ \{1,2,3\}\times\{1,2,3\} = \left\{ \begin{array}{ccc} (1,1), & (1,2), & (1,3), \\ (2,1), & (2,2), & (2,3), \\ (3,1), & (3,2), & (3,3)\phantom{,} \end{array} \right\}. $$ Suppose the event $A$ is that either a $1$ or a $2$ appears on the first trial, and the event $B$ is that a $1$ appears on the second trial. Then $$ A = \left\{ \begin{array}{ccc} (1,1), & (1,2), & (1,3), \\ (2,1), & (2,2), & (2,3)\phantom{,} \end{array} \right\} \text{ and } B= \left\{ \begin{array}{ccc} (1,1), \\ (2,1), \\ (3,1)\phantom{,} \end{array} \right\}. $$ The intersection is just the set of outcomes in both $A$ and $B$, meaning both events occur: $$ A\cap B = \left\{ \begin{array}{c} (1,1), \\ (2,1)\phantom{,} \end{array} \right\}. $$