Probability · Higher Level
Sets & Probability
From the language of sets through to independence — the full HL toolkit, one tap at a time.
Section 1
The Language of Sets
Before we go near probability, you have to know every symbol cold. They turn up in nearly every question.
Set = a group with a common trait.
Must learn — the symbol vocabulary
$\cup$
union = together = or
$\cap$
intersection = common
$\varnothing$ or $\{\,\}$
null = no elements
$A'$
complement = outside $A$
$\#$
cardinal = amount of elements
$\mathcal{U}$
universal = all possible elements
$\in$
element = part of a set
$\subset$
subset = set in a set
(i) Reading the symbols out loud
When you see one of these on the page, say it to yourself in words. That's the first translation step.
$A \cup B$ (A or B — together)
$A \cap B$ (A and B — common to both)
$\#A$ (how many elements are in A)
$A'$ (everything outside A)
$A \cap B'$ (in A AND outside B)
Try Me
Read $P(A' \cap B)$ out loud — what is it asking for in plain English?
Translate piece by piece: $A'$ first, then $\cap$, then $B$.
$A'$ = outside $A$
$\cap$ = and (common to both)
$B$ = inside $B$
Stitch them together.
Probability that an element is outside $A$ AND inside $B$.
Section 2
Reading a Venn Diagram
Every Venn question starts the same way: read the four regions, add them up to get the total — that's your $\#\mathcal{U}$.
Given: $\#A = 14$, $\#B = 15$.
First the total. Add every number you see — the three regions inside the circles and the bit outside.
Total = $\#\mathcal{U}$ = $5 + 9 + 6 + 3$ = $23$
(i) Find $P(A)$
$P(A)$ means "the probability of landing inside $A$". Take the whole $A$ circle over the total.
$\#A = 5 + 9 = 14$
$P(A) = \dfrac{14}{23}$
(ii) Find $P(A \cap B)$
The intersection is the overlap — common to both circles. Just the middle number.
$P(A \cap B) = \dfrac{9}{23}$
(iii) Find $P(A \cup B)$
The union is everything inside either circle. Add the three numbers that sit inside the two circles.
$5 + 9 + 6 = 20$
$P(A \cup B) = \dfrac{20}{23}$
(iv) Find $P(A')$
$A'$ is everything outside $A$ — that's the bit of $B$ on its own, plus the 3 sitting outside both circles.
Outside $A$ = $6 + 3 = 9$
$P(A') = \dfrac{9}{23}$
(v) Find $P(A \cap B')$
In $A$ and outside $B$ — that's the slice of $A$ that doesn't overlap $B$.
$P(A \cap B') = \dfrac{5}{23}$
Try Me
A new Venn: 7 in $A$ only, 4 in the overlap, 8 in $B$ only, 6 outside. Find $P(A)$ and $P(A \cup B)$.
Total first, then read each one off the diagram.
Total = $7 + 4 + 8 + 6 = 25$
$\#A = 7 + 4 = 11$
$P(A) = \dfrac{11}{25}$
$\#(A \cup B) = 7 + 4 + 8 = 19$
$P(A \cup B) = \dfrac{19}{25}$
$P(A) = \dfrac{11}{25}, \quad P(A \cup B) = \dfrac{19}{25}$
Try Me
Using the same Venn as above (7, 4, 8, 6), find $P(B')$ and $P(A' \cap B')$.
$B'$ is everything outside $B$. $A' \cap B'$ is outside both — only one region fits that.
Outside $B$ = $7 + 6 = 13$
$P(B') = \dfrac{13}{25}$
Outside $A$ AND outside $B$ = just the 6 sitting outside both circles.
$P(A' \cap B') = \dfrac{6}{25}$
$P(B') = \dfrac{13}{25}, \quad P(A' \cap B') = \dfrac{6}{25}$
Section 3
The Union Formula
There's a slick formula for $P(A \cup B)$. It earns its keep when you're given the probabilities or cardinals but the Venn isn't drawn.
Must learn
$P(A \cup B) \;=\; P(A) + P(B) - P(A \cap B)$
Why subtract the intersection? Because if you just add $P(A)$ and $P(B)$, the overlap gets counted twice. You're paying the toll twice on the same bridge. Take one off.
(i) Check it on the last Venn
From §2 we had $P(A) = \dfrac{14}{23}$, $P(B) = \dfrac{15}{23}$, $P(A \cap B) = \dfrac{9}{23}$.
$P(A \cup B) = \dfrac{14}{23} + \dfrac{15}{23} - \dfrac{9}{23}$
$= \dfrac{14 + 15 - 9}{23}$
$= \dfrac{20}{23}$ ✓ matches the direct count.
(ii) A typical exam-style use
$P(A) = 0.6$, $P(B) = 0.5$, $P(A \cap B) = 0.2$. Find $P(A \cup B)$.
$P(A \cup B) = 0.6 + 0.5 - 0.2$
$P(A \cup B) = 0.9$
Try Me
$P(A) = 0.45$, $P(B) = 0.30$, $P(A \cap B) = 0.12$. Find $P(A \cup B)$.
Plug in the formula. Watch the minus sign.
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$= 0.45 + 0.30 - 0.12$
$= 0.63$
$P(A \cup B) = 0.63$
Try Me
$P(A) = \dfrac{1}{2}$, $P(B) = \dfrac{2}{5}$, $P(A \cup B) = \dfrac{3}{4}$. Find $P(A \cap B)$.
Rearrange the formula: solve for the intersection. Common denominator is 20.
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$\dfrac{3}{4} = \dfrac{1}{2} + \dfrac{2}{5} - P(A \cap B)$
$\dfrac{15}{20} = \dfrac{10}{20} + \dfrac{8}{20} - P(A \cap B)$
$\dfrac{15}{20} = \dfrac{18}{20} - P(A \cap B)$
$P(A \cap B) = \dfrac{18}{20} - \dfrac{15}{20}$
$P(A \cap B) = \dfrac{3}{20}$
$P(A \cap B) = \dfrac{3}{20}$
Section 4
Conditional Probability $P(A \mid B)$
Some questions change the universe on you. "Given that $B$ has happened…" shrinks your world down to just the $B$ circle. Everything outside $B$ is gone.
$P(A \mid B)$ = probability of $A$ given $B$ has occurred.
(i) Pick a soccer player at random — what's the chance they also play hurling?
The wording is: play hurling given you play soccer. So $B$ has happened — we're inside the soccer circle only.
$\#B = 9 + 6 = 15$ (the new universe)
Out of those 15 soccer players, how many ALSO play hurling? The overlap: 9.
$P(A \mid B) = \dfrac{9}{15}$
$P(A \mid B) = \dfrac{3}{5}$
Must learn — the conditional formula
$P(A \mid B) \;=\; \dfrac{P(A \cap B)}{P(B)}$
"What's common, over the new universe."
A warning on notation: at Junior Cert, $A \mid B$ sometimes meant "$A$ less $B$" — the set difference. Forget that. At Leaving Cert HL, $A \mid B$ means given $B$.
(ii) The reverse — $P(B \mid A)$
Same Venn. Pick a hurler — what's the chance they also play soccer? Now $A$ has happened, so we live inside the hurling circle.
$\#A = 5 + 9 = 14$ (new universe)
$P(B \mid A) = \dfrac{9}{14}$
Notice $P(A \mid B) = \dfrac{9}{15}$ and $P(B \mid A) = \dfrac{9}{14}$ are NOT the same. The denominator changes depending on which event you're conditioning on.
Try Me
Venn: 8 in $A$ only, 6 in the overlap, 10 in $B$ only, 4 outside. Find $P(A \mid B)$.
$B$ has happened — find $\#B$ first. Then ask: how many of those are also in $A$?
$\#B = 6 + 10 = 16$ (new universe)
In $A$ as well = overlap = $6$
$P(A \mid B) = \dfrac{6}{16}$
$P(A \mid B) = \dfrac{3}{8}$
$P(A \mid B) = \dfrac{3}{8}$
Try Me
$P(A \cap B) = 0.18$ and $P(B) = 0.4$. Find $P(A \mid B)$.
Straight into the formula.
$P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$
$= \dfrac{0.18}{0.4}$
$= 0.45$
$P(A \mid B) = 0.45$
Section 5
Independence
Are hurling and soccer connected? Compare the plain $P(A)$ with the conditional $P(A \mid B)$.
$P(A) = \dfrac{14}{23} \approx 0.609$
$P(A \mid B) = \dfrac{9}{15} = 0.6$
These are close, but not equal: $\dfrac{14}{23}$ ≠ $\dfrac{9}{15}$. So knowing $B$ does change the probability of $A$.
These events are NOT independent, since $P(A \mid B)$ ≠ $P(A)$.
Must learn — Independent
One has no effect on the other. Either of these tests works:
$P(A \mid B) = P(A)$
$P(A) \cdot P(B) = P(A \cap B)$
In practice the multiplication test, $P(A) \cdot P(B) = P(A \cap B)$, is usually the cleanest — no division, no awkward fractions.
(i) Testing independence
$P(A) = 0.4$, $P(B) = 0.5$, $P(A \cap B) = 0.2$. Are $A$ and $B$ independent?
$P(A) \cdot P(B) = 0.4 \times 0.5 = 0.2$
$P(A \cap B) = 0.2$ (given)
$P(A) \cdot P(B) = P(A \cap B)$ ✓ Yes, independent.
(ii) A pair that aren't
$P(A) = 0.5$, $P(B) = 0.3$, $P(A \cap B) = 0.2$. Independent?
$P(A) \cdot P(B) = 0.5 \times 0.3 = 0.15$
$P(A \cap B) = 0.2$
$0.15$ ≠ $0.2$ — NOT independent.
Try Me
$P(A) = \dfrac{1}{3}$, $P(B) = \dfrac{1}{4}$, $P(A \cap B) = \dfrac{1}{12}$. Are $A$ and $B$ independent?
Multiply $P(A)$ and $P(B)$ — does it equal $P(A \cap B)$?
$P(A) \cdot P(B) = \dfrac{1}{3} \times \dfrac{1}{4} = \dfrac{1}{12}$
$P(A \cap B) = \dfrac{1}{12}$ (given)
Yes — independent.
Independent. $P(A) \cdot P(B) = P(A \cap B) = \dfrac{1}{12}$.
Try Me
$P(A) = 0.6$, $P(B) = 0.5$, $P(A \cap B) = 0.4$. Are $A$ and $B$ independent?
Same test. Multiply and compare.
$P(A) \cdot P(B) = 0.6 \times 0.5 = 0.3$
$P(A \cap B) = 0.4$ (given)
$0.3$ ≠ $0.4$
NOT independent.
NOT independent. $P(A) \cdot P(B) = 0.3$ but $P(A \cap B) = 0.4$.
Section 6
Mutually Exclusive & Exhaustive
Two more bits of vocabulary that show up constantly. Both describe the shape of the Venn diagram.
Must learn — Mutually exclusive
Nothing in common. The two circles don't overlap.
$P(A \cap B) = 0$
Must learn — Exhaustive
No other options — only $A$ or $B$ can happen. Nothing sits outside the two circles.
$P(A \cup B) = 1$
(i) Putting both together
$A$ and $B$ are mutually exclusive and exhaustive. $P(B) = 0.4$. Find $P(A)$.
Circles separated (mutually exclusive) & nothing outside (exhaustive).
Mutually exclusive means no overlap to worry about. Exhaustive means everything sits inside the two circles — outside is empty.
$P(A) + P(B) = 1$ (exhaustive, no overlap)
$P(A) = 1 - 0.4$
$P(A) = 0.6$
(ii) Independent — fractions
$P(A) = \dfrac{1}{3}$, $P(B) = \dfrac{1}{2}$, and $A$ and $B$ are independent. Find $P(A \cap B)$ and $P(A \cup B)$.
$P(A \cap B)$ first — use the multiplication test for independence.
$P(A \cap B) = P(A) \cdot P(B)$
$= \dfrac{1}{3} \times \dfrac{1}{2}$
$P(A \cap B) = \dfrac{1}{6}$
Then $P(A \cup B)$ using the union formula. Common denominator is 6.
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$= \dfrac{2}{6} + \dfrac{3}{6} - \dfrac{1}{6}$
$P(A \cup B) = \dfrac{4}{6} = \dfrac{2}{3}$
Try Me
$A$ and $B$ are mutually exclusive. $P(A) = 0.3$ and $P(B) = 0.45$. Find $P(A \cup B)$.
If mutually exclusive, what's the intersection? Then use the union formula.
Mutually exclusive $\Rightarrow$ $P(A \cap B) = 0$
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$= 0.3 + 0.45 - 0$
$= 0.75$
$P(A \cup B) = 0.75$
Try Me
$A$ and $B$ are independent. $P(A) = \dfrac{2}{5}$ and $P(B) = \dfrac{1}{4}$. Find $P(A \cup B)$.
Independent gives you $P(A \cap B)$. Common denominator 20.
Independent $\Rightarrow$ $P(A \cap B) = P(A) \cdot P(B)$
$= \dfrac{2}{5} \times \dfrac{1}{4} = \dfrac{2}{20} = \dfrac{1}{10}$
$P(A \cup B) = \dfrac{2}{5} + \dfrac{1}{4} - \dfrac{1}{10}$
$= \dfrac{8}{20} + \dfrac{5}{20} - \dfrac{2}{20}$
$= \dfrac{11}{20}$
$P(A \cup B) = \dfrac{11}{20}$
Section 7
The Full Six-Part Question
This is the type of question that comes up every year. One Venn, one set of probabilities, six things to find. We'll work all six in order.
Find: (i) $P(A \cup B)$ (ii) $P(A \mid B)$ (iii) $P(A' \cap B')$ (iv) $x$ (v) $P(B \mid A)$ (vi) Independent?
First, read the three numbers off the Venn. $P(A) = 0.2 + 0.3 = 0.5$. $P(B) = 0.3 + 0.1 = 0.4$. $P(A \cap B) = 0.3$. Keep these on the side.
(i) $P(A \cup B)$
Two ways. Either add the three regions inside the circles, or use the formula. Both should agree.
$P(A \cup B) = 0.2 + 0.3 + 0.1 = 0.6$ (direct count)
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$= 0.5 + 0.4 - 0.3$
$P(A \cup B) = 0.6$ ✓
(ii) $P(A \mid B)$
$P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$
$= \dfrac{0.3}{0.4}$
$P(A \mid B) = 0.75$
(iii) $P(A' \cap B')$
Outside $A$ AND outside $B$ — that's the region sitting outside both circles. That's just $x$ on the diagram.
$P(A' \cap B') = x$
But we haven't found $x$ yet — that's part (iv). Let's pause this and finish $x$ first.
(iv) Find $x$
All four regions add to 1. Everyone has to land somewhere.
$0.2 + 0.3 + 0.1 + x = 1$
$0.6 + x = 1$
$x = 0.4$
So back to (iii): $P(A' \cap B') = x = \mathbf{0.4}$.
(v) $P(B \mid A)$
Same formula, the other way round.
$P(B \mid A) = \dfrac{P(B \cap A)}{P(A)}$
$= \dfrac{0.3}{0.5}$
$P(B \mid A) = 0.6$
(vi) Independent?
Two tests — use whichever is easier. The multiplication one is usually cleanest.
$P(A) \cdot P(B) = 0.5 \times 0.4 = 0.2$
$P(A \cap B) = 0.3$ (from the Venn)
$0.2$ ≠ $0.3$ — they're not equal.
NOT independent.
Cross-check with the other test: $P(A \mid B) = 0.75$ and $P(A) = 0.5$. Different, so not independent. Both tests agree, as they always must.
Try Me
New Venn: $0.15$ in $A$ only, $0.25$ in the overlap, $0.35$ in $B$ only, $y$ outside. Find (a) $y$, (b) $P(A \cup B)$, (c) $P(A \mid B)$.
$y$ first (everything must total 1). Then read $P(A)$, $P(B)$, $P(A \cap B)$ off the diagram and plug in.
(a) $0.15 + 0.25 + 0.35 + y = 1$
$0.75 + y = 1 \;\Rightarrow\; y = 0.25$
(b) $P(A \cup B) = 0.15 + 0.25 + 0.35 = 0.75$
(c) $P(B) = 0.25 + 0.35 = 0.6$
$P(A \cap B) = 0.25$
$P(A \mid B) = \dfrac{0.25}{0.6}$
$= \dfrac{5}{12} \approx 0.417$
$y = 0.25, \quad P(A \cup B) = 0.75, \quad P(A \mid B) = \dfrac{5}{12}$
Try Me
Using the same Venn (0.15, 0.25, 0.35, 0.25), are $A$ and $B$ independent?
$P(A) \cdot P(B)$ vs $P(A \cap B)$.
$P(A) = 0.15 + 0.25 = 0.4$
$P(B) = 0.25 + 0.35 = 0.6$
$P(A) \cdot P(B) = 0.4 \times 0.6 = 0.24$
$P(A \cap B) = 0.25$
$0.24$ ≠ $0.25$
NOT independent (but only just).
NOT independent. $P(A) \cdot P(B) = 0.24$ ≠ $0.25 = P(A \cap B)$.
Section 8
Find the Missing Probability
Last type — they hand you most of the story and ask you to dig out the missing piece using independence as a tool.
(i) Given $P(A)$ and $P(A \cup B)$ — find $P(B)$
$A$ and $B$ are independent. $P(A) = 0.2$ and $P(A \cup B) = 0.5$. Find $P(B)$.
The trap: students reach for $P(A \mid B) = P(A)$ or $P(B \mid A) = P(B)$ first. Watch where that leads.
$P(B \mid A) = P(B)$ (start of wrong attempt)
$\dfrac{P(B \cap A)}{P(A)} = P(B)$
$\dfrac{0.2 \cdot P(B)}{0.2} = P(B)$ (used $P(A \cap B) = P(A)P(B)$)
$P(B) = P(B)$ (useless — true but tells you nothing)
The conditional rule on its own just churns the wheel. The right move is to use the other version of independence, then plug straight into the union formula.
Correct method. Let $P(B) = x$.
Independent $\Rightarrow$ $P(A \cap B) = P(A) \cdot P(B) = 0.2x$
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$0.5 = 0.2 + x - 0.2x$
$0.5 - 0.2 = x - 0.2x$
$0.3 = 0.8x$
$x = \dfrac{0.3}{0.8}$
$P(B) = \dfrac{3}{8} = 0.375$
Lesson: when independence is given, almost always use $P(A \cap B) = P(A) \cdot P(B)$ as a substitution into the union formula. That's the workhorse.
Try Me
$A$ and $B$ are independent. $P(A) = 0.4$ and $P(A \cup B) = 0.7$. Find $P(B)$.
Let $P(B) = x$. Use the union formula with $P(A \cap B) = P(A)P(B) = 0.4x$.
Let $P(B) = x$. Independent $\Rightarrow$ $P(A \cap B) = 0.4x$
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$0.7 = 0.4 + x - 0.4x$
$0.3 = 0.6x$
$P(B) = 0.5$
$P(B) = 0.5$
Try Me
$A$ and $B$ are independent. $P(B) = \dfrac{1}{3}$ and $P(A \cup B) = \dfrac{2}{3}$. Find $P(A)$.
Let $P(A) = x$. Same play — sub into the union formula.
Let $P(A) = x$. Independent $\Rightarrow$ $P(A \cap B) = x \cdot \dfrac{1}{3} = \dfrac{x}{3}$
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$\dfrac{2}{3} = x + \dfrac{1}{3} - \dfrac{x}{3}$
$\dfrac{2}{3} - \dfrac{1}{3} = x - \dfrac{x}{3}$
$\dfrac{1}{3} = \dfrac{2x}{3}$
$1 = 2x$
$P(A) = \dfrac{1}{2}$
$P(A) = \dfrac{1}{2}$
Try Me
$A$ and $B$ are mutually exclusive and exhaustive. $P(A) = \dfrac{3}{5}$. Find $P(B)$ and $P(A \cap B)$.
Mut exc: no overlap. Exhaustive: nothing outside.
Mutually exclusive $\Rightarrow$ $P(A \cap B) = 0$
Exhaustive $\Rightarrow$ $P(A \cup B) = 1$
$P(A) + P(B) - 0 = 1$
$\dfrac{3}{5} + P(B) = 1$
$P(B) = \dfrac{2}{5}, \quad P(A \cap B) = 0$
$P(B) = \dfrac{2}{5}, \quad P(A \cap B) = 0$
Sets & Probability — done.
Eight sections, sixteen try-mes. You've got every Venn-style HL question covered now.