L03 - Mutually and Non-Mutually Exclusive Events

Mutually and Non-Mutually Exclusive Events

Learning Intentions

• I can apply the addition rule.
• I can use the addition rule.

Definition of Mutually Exclusive Events

• Mutually exclusive events mean that both outcomes are exclusive to each other, meaning there is no overlap.
  • Mutually = both
  • Exclusive = excluding, separate
• $A\cap B=\varnothing =\{\}$

Example of Mutually Exclusive Events

• Let’s take the sample space of rolling a standard six-sided die.
  • If outcome A = odd numbers and
  • outcome B = even numbers,
  • then $A\cap B=\{\}$ (mutually exclusive, see the definition.)

Definition of Non-Mutually Exclusive Events

• Non-mutually exclusive events mean that both outcomes are not exclusive to each other, meaning there is no overlap.
• $A\cap B\ne \varnothing$

Example of Mutually Exclusive Events

• Let’s take the sample space of rolling a standard six-sided die.
  • If outcome A = odd numbers and
  • outcome B = even numbers,
  • then $A\cap B=\{\}$ (mutually exclusive, see the definition.)

The Addition Rule

The Addition Rule

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

• This works by adding the probability of the two outcomes (which overlap) and subtracting the overlap.
• $P(A\cap B)$ is ignored in a mutually exclusive scenario.

Standard Application of The Addition Rule (1)

• A card is selected from a standard deck of 52 playing cards (no jokers). Find $P(A\cup B)$.
  • A = card of diamonds
  • B = jack card
• $P(A)=\frac{1}{4}$ , $P(B)=\frac{1}{13}$, $P(A\cap B)=\frac{1}{52}$
• Apply the addition rule to find $P(A\cup B)$.

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

$P(A\cup B)=\frac{1}{4}+\frac{1}{13}-\frac{1}{52}=\frac{13+4-1}{52}$

Standard Application of The Addition Rule (2)

• Two events, A and B, are such that
  • $P(A)=0.4$
  • $P(B)=0.8$
  • $P(A\cup B)=0.85$
• Find (a) $P(A\cap B)$ and (b) $P(A^{\prime }\cap B^{\prime })$.
• (a) $P(A\cap B)$ = 0.35
  • $P(A)+P(B)=0.4+0.8=1.2$
  • $P(A\cup B)=1.2\boxed{P(A)\; +\; P(B)}$
  • $P(A\cap B)=1.2-0.85$
  • $\therefore P(A\cap B)=0.35$
• (b) $P(A^{\prime }\cap B^{\prime })=0.15$
  • $1-0.85=0.15$ (anything not A and not B, complement!)