# 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}

## Answer and Explanation:

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}.

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from Statistics 101: Principles of Statistics

Chapter 4 / Lesson 10