Exercises — Mutually exclusive events — addition rule

Can both happen at once? A single roll gives one face; it cannot be a $3$ and a $6$ simultaneously. So $A\cap B=\varnothing$ — they are mutually exclusive. $P(A)=\frac{1}{6}$, $P(B)=\frac{1}{6}$. With no overlap: $P(A\cup B)=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}.$

Can both happen at once? Yes — the King of Hearts is a heart and a King. So $A\cap B=\{\text{King of Hearts}\}\ne \varnothing$. Therefore $A$ and $B$ are not mutually exclusive. (You would need the general rule, subtracting $P(A\cap B)=\frac{1}{52}$, to combine them.)

These are three single faces, no two of which can occur together — pairwise mutually exclusive. For pairwise exclusive events you just add: $P=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}.$

Can both happen? Yes — the Ace of Spades. Not exclusive, so use the general rule. $P(\text{Ace})=\frac{4}{52}$, $P(\text{Spade})=\frac{13}{52}$, overlap $P(\text{Ace}\cap \text{Spade})=\frac{1}{52}$. $P=\frac{4}{52}+\frac{13}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}.$ Check by counting: 4 Aces + 13 Spades − 1 double-counted (Ace of Spades) = $16$ favourable cards. ✓

Colours are exclusive (a ball is exactly one colour), so add: $P(\text{red}\cup \text{green})=0.4+0.25=\mathrm{0.65.}$ For "not blue", use the complement rule $P(A^{\prime })=1-P(A)$: $P(\text{not blue})=1-0.35=\mathrm{0.65.}$ (Same number — because "not blue" is "red or green" here, since the three colours are exhaustive.)

Rearrange the general rule to find the overlap: $P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.45+0.35-0.8=0.$ The overlap is $0$, so $A\cap B=\varnothing$ — they are mutually exclusive. (When plain addition of $P(A)$ and $P(B)$ exactly reproduces $P(A\cup B)$, there was nothing to subtract.)

$P(A\cap B)=0.5+0.4-0.7=\mathrm{0.2.}$ The overlap is $0.2\ne 0$, so they are not mutually exclusive — they share $20\%$ of the probability.

If they were exclusive, $P(A\cup B)=0.55+0.6=1.15>1$ — impossible, since probability caps at $1$. So they cannot be mutually exclusive. The union is at most $1$, so from $P(A\cap B)=P(A)+P(B)-P(A\cup B)$ the overlap is smallest when $P(A\cup B)$ is largest ($=1$): $P(A\cap B{)}_{\min }=0.55+0.6-1=\mathrm{0.15.}$

Let $F$ = football, $C$ = cricket. Given $P(F)=0.4$, $P(C)=0.3$, $P(F\cap C)=0.1$. (a) At least one = union. General rule (they overlap!): $P(F\cup C)=0.4+0.3-0.1=\mathrm{0.6.}$ (b) "Neither" is the complement of "at least one": $P(\text{neither})=1-P(F\cup C)=1-0.6=\mathrm{0.4.}$ (c) "Football only" = football minus the shared part: $P(F\text{ only})=P(F)-P(F\cap C)=0.4-0.1=\mathrm{0.3.}$

Independent means "$A$ tells you nothing about $B$", which fixes the intersection via the multiplication rule: $P(A\cap B)=P(A)P(B)=0.5\times 0.4=\mathrm{0.2.}$ Then the general addition rule gives $P(A\cup B)=0.5+0.4-0.2=\mathrm{0.7.}$ Mutually exclusive would instead force $P(A\cap B)=0$, so $P(A\cup B)=0.5+0.4=\mathrm{0.9.}$ Different answers — because independence and exclusivity are different constraints on the overlap (one makes it $P(A)P(B)$, the other makes it $0$).

Exhaustive & exclusive means all four sum to $1$: $0.3+0.25+P(\text{green})+P(\text{yellow})=1\Rightarrow P(\text{green})+P(\text{yellow})=\mathrm{0.45.}$ Substitute $P(\text{green})=2P(\text{yellow})$: $2P(\text{yellow})+P(\text{yellow})=0.45\Rightarrow 3P(\text{yellow})=0.45\Rightarrow P(\text{yellow})=\mathrm{0.15.}$ So $P(\text{green})=0.30$. Check: $0.3+0.25+0.30+0.15=1$. ✓ Red and yellow are exclusive, so add: $P(\text{red}\cup \text{yellow})=0.3+0.15=\mathrm{0.45.}$

Count outcomes ($n(S)=30$):

• Multiples of $3$: $3,6,\dots ,30$ → $10$ numbers, $P(A)=\frac{10}{30}$.
• Multiples of $5$: $5,10,\dots ,30$ → $6$ numbers, $P(B)=\frac{6}{30}$.
• Both = multiples of $15$: $15,30$ → $2$ numbers, $P(A\cap B)=\frac{2}{30}$.

They overlap (e.g. $15$), so use the general rule: $P(A\cup B)=\frac{10}{30}+\frac{6}{30}-\frac{2}{30}=\frac{14}{30}=\frac{7}{15}.$ "Neither" is the complement: $P(\text{neither})=1-\frac{14}{30}=\frac{16}{30}=\frac{8}{15}.$

The six faces are mutually exclusive and exhaustive, so their probabilities sum to $1$: ${\sum }_{k=1}^{6}ck=c(1+2+3+4+5+6)=21c=1\Rightarrow c=\frac{1}{21}.$ Primes are $\{2,3,5\}$ — distinct faces, so mutually exclusive; add their probabilities: $P(\text{prime})=\frac{2}{21}+\frac{3}{21}+\frac{5}{21}=\frac{10}{21}.$