Q.Given $P(A) = \dfrac{3}{5}$ and $P(B) = \dfrac{1}{5}$, find $P(A \text{ or } B)$, if $A$ and $B$ are mutually exclusive events. OR A coin is tossed twice. What is the probability that at least one tail occurs?

Events A and B are mutually exclusive if they cannot occur simultaneously, i.e. $A\cap B=\varnothing$.

In a random experiment with sample space S, events $A,B\subseteq S$ are mutually exclusive if $A\cap B=\varnothing$ — no outcome belongs to both events.

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$P(A\cup B)=0.8$, so $P(A'\cap B')=1-0.8=0.2$.

Since $A,B$ are mutually exclusive, $P(A\cap B)=0$, hence

$P(A\cup B)=P(A)+P(B)=0.5+0.3=0.8.$

By De Morgan's law $A'\cap B'=(A\cup B)'$, so …

Mutually exclusive events cannot occur together, so $P(E \cap F) = 0$ and the union rule simplifies to a plain sum.

Since $E$ and $F$ are mutually exclusive, $P(E \cap F) = 0$.

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Mutually exclusive events don't overlap, so their probabilities simply add: $3/5+1/5=4/5$.

For mutually exclusive events $A$ and $B$: $P(A\cap B)=0$, so

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This is exactly the addition theorem of probability restricted to mutually exclusive events, and it is correct — the statement is True.

The general addition theorem of probability states:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Mutually exclusive events are events that cannot occur at the same time, meaning $A \cap B = \emptyset$, so $P(A \cap B) = 0$.

Substituting into the general formula: …

For mutually exclusive events, $P(A\cup B)=P(A)+P(B)$.

Since $A$ and $B$ are mutually exclusive (they cannot both occur), $P(A\cap B)=0$, so: …

Mutually exclusive events have $A \cap B = \phi$.

Two events $A$ and $B$ are called mutually exclusive if they can never occur at the same time, i.e. they share no common outcomes.

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Mutually exclusive means the two events share no common outcome, so their intersection is the empty set $\phi$.

Two events $A$ and $B$ are called mutually exclusive if they cannot occur simultaneously — that is, there is no outcome common to both.

Since there is no outcome common to $A$ and $B$, their intersection (the set of common outcomes) is empty: …

Mutually exclusive events cannot occur together, so their intersection is the empty event with probability $0$.

By definition, two events $A$ and $B$ are mutually exclusive if they cannot occur simultaneously, i.e. $A\cap B=\varnothing$.

Hence $P(A\cap B) = P(\varnothing) = 0$.

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