# Let A and B be two events such that P (A) = 0.46 and P (B) = 0.07. (a) Determine P (A union B),...

## Question:

Let {eq}A {/eq} and {eq}B {/eq} be two events such that {eq}P (A) = 0.46 {/eq} and {eq}P (B) = 0.07 {/eq}.

(a) Determine {eq}P (A \cup B), {/eq} given that {eq}A {/eq} and {eq}B {/eq} are independent.

(B) Determine {eq}P (A \cup B) {/eq} given that {eq}A {/eq} and {eq}B {/eq} are mutually exclusive.

## Independent events and mutually exclusive events

Two events are said to be independent if the probability of happening of one event doesn't affect the probability of happening of the other event.

{eq}P\left( {A \cap B} \right) = P\left( A \right)P\left( B \right) {/eq}, given that the events are independent.

If the events are mutually exclusive, {eq}P\left( {A \cap B} \right) = 0. {/eq}

Given

{eq}P\left( A \right) = 0.46,P\left( B \right) = 0.07 {/eq}.

We know that given independent events, {eq}P\left( {A \cap B} \right) = P\left( A \right)P\left( B \right) {/eq}.

{eq}\begin{align*} P\left( {A \cup B} \right) &= P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\\ P\left( {A \cup B} \right) &= P\left( A \right) + P\left( B \right) - P\left( A \right)P\left( B \right)\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= 0.46 + 0.07 - 0.46*0.07\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= 0.497800 \end{align*} {/eq}.

Given mutually exclusive events, {eq}P\left( {A \cap B} \right) = 0 {/eq}.

{eq}\begin{align*} P\left( {A \cup B} \right) &= P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\\ P\left( {A \cup B} \right) &= P\left( A \right) + P\left( B \right) - 0\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= 0.46 + 0.07\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= 0.53 \end{align*} {/eq}.