Q.If $A$ and $B$ are mutually exclusive events with $P(A) = \dfrac{3}{8}$ and $P(B) = \dfrac{1}{8}$, then find

Among 1 to 20, there are 6 multiples of 3, 5 multiples of 4, and 1 multiple of both (12); by inclusion-exclusion that's 10 numbers, giving probability 10/20 = 1/2.

Multiples of 3 in $\{1,\ldots,20\}$: 3,6,9,12,15,18 — 6 numbers.

Multiples of 4: 4,8,12,16,20 — 5 numbers.

Multiples of both 3 and 4 (i.e. of 12): 12 — 1 number.

By inclusion-exclusion, numbers divisible by 3 or 4 $=6+5-1=10$.

$P(\text{divisible by 3 or 4})=\dfrac{10}{20}=\dfrac{1}{2}$.

"Exactly one occurs" splits into two mutually exclusive cases: A but not B, or B but not A.

The event "exactly one of A, B occurs" is $(A\cap\bar B)\cup(\bar A\cap B)$. These two events are mutually exclusive (they can't both happen), so by the addition rule,

$P(\text{exactly one})=P(A\cap\bar B)+P(\bar A\cap B).$

$P(A\cap B)=\dfrac15$.

Since $P(\bar A)=\dfrac{3}{10}$, we get $P(A)=1-\dfrac{3}{10}=\dfrac{7}{10}$.

Event A splits into the disjoint parts $A\cap B$ and $A\cap\bar B$, so

$P(A)=P(A\cap B)+P(A\cap\bar B).$

Given $P(A\cap\bar B)=\dfrac12$:

$\dfrac{7}{10}=P(A\cap B)+\dfrac12\ \Rightarrow\ P(A\cap B)=\dfrac{7}{10}-\dfrac{5}{10}=\dfrac{2}{10}=\dfrac15.$

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

For any two events: $P(A\cup B) = P(A)+P(B)-P(A\cap B)$.

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

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

Since $A$ and $B$ share no outcomes, every outcome of $A$ automatically lies outside $B$, so $A\cap\bar B = A$.

Mutually exclusive means $A\cap B=\emptyset$ — no outcome belongs to both. So every outcome in $A$ is necessarily NOT in $B$, meaning $A\subseteq \bar B$.

Therefore $A\cap\bar B = A$, and:

$P(A\cap\bar B) = P(A) = 0.5$

Since $A\cap B=\emptyset$, every outcome of $B$ lies outside $A$, so $\bar A\cap B = B$.

Mutually exclusive means $A\cap B=\emptyset$, so $B\subseteq\bar A$ (no outcome of $B$ is in $A$).

Therefore $\bar A\cap B = B$, and:

$P(\bar A\cap B) = P(B) = 0.3$

'Exactly one occurs' splits into two disjoint cases whose probabilities simply add.

'Exactly one of $A$ and $B$ occurs' means either ($A$ occurs and $B$ does not) or ($B$ occurs and $A$ does not). In set notation these are $A\cap\bar B$ (A but not B) and $\bar A\cap B$ (B but not A). These two events are mutually exclusive (they cannot both happen at once), so by the addition rule for disjoint events:

$P(\text{exactly one})=P(A\cap\bar B)+P(\bar A\cap B).$

This is distinct from $P(A)+P(B)-P(A\cap B)$, which is $P(A\cup B)$ (at least one occurs, including both).

Count multiples of 3 (6 numbers), multiples of 4 (5 numbers), subtract the overlap — multiples of 12 (1 number) — to get 10 favourable outcomes out of 20, so the probability is $1/2$.

Multiples of 3 in $\{1,\ldots,20\}$: $3,6,9,12,15,18$ — that's $\lfloor 20/3\rfloor = 6$ numbers.

Multiples of 4: $4,8,12,16,20$ — that's $\lfloor 20/4\rfloor = 5$ numbers.

Multiples of both 3 and 4, i.e. multiples of 12: $12$ — that's $\lfloor 20/12\rfloor = 1$ number.

By inclusion–exclusion, numbers divisible by 3 or 4 $= 6+5-1 = 10$.

Probability $= \dfrac{10}{20} = \dfrac{1}{2}$.