Probability basics — sample space, events, P(E) = favourable - total: Worked examples

This page is a practice arena. The parent note built the machinery; here we throw every kind of case at it — normal counts, the two edge cases (probability $0$ and probability $1$ ), the trap of unequal outcomes, "or"/"and" overlaps, a word problem, and an exam twist — and solve each fully.

Before a single number: let us agree on the only three quantities we ever compute.

Recall The whole machine in one line

We run an experiment. We list every equally-likely outcome — that set is the sample space $S$ , and its size is written $n (S)$ . We circle the outcomes we care about — that subset is the event $E$ , size $n (E)$ . Then $P (E) = \frac{n ( E )}{n ( S )} .$ $n$ just means "number of things in". Nothing else happens on this page except counting and dividing.

The scenario matrix

Probability problems, once you strip the story away, fall into a small number of case classes. This table lists them all. Each worked example below is tagged with the cell it fills, so by the end you will have seen every row.

#	Case class	What makes it tricky	Example that covers it
A	Ordinary count, one experiment	nothing — the base case	Ex 1
B	Degenerate: impossible event ( $P = 0$ )	the event is the empty set $\emptyset$	Ex 2
C	Degenerate: sure event ( $P = 1$ )	the event is all of $S$	Ex 2
D	Outcomes not equally likely	naive $n (E) / n (S)$ is illegal	Ex 3
E	"OR" — union with overlap	double-counting shared outcomes	Ex 4
F	"AND" — intersection	must keep only shared outcomes	Ex 4
G	Complement — "at least one"	count the easy opposite instead	Ex 5
H	Ordered vs unordered sample space	choosing the right $S$	Ex 6
I	Word problem (real-world)	translating words into $S$ and $E$	Ex 7
J	Exam twist — combine several rules	staged reasoning	Ex 8

Two visual anchors first, then the examples.

Figure — Probability basics — sample space, events, P(E) = favourable - total

Read the picture as: a big box is the whole sample space $S$ ; the coloured blob inside is the event $E$ . Probability is literally how much of the box the blob covers. An empty blob covers nothing → $P = 0$ . A blob filling the box → $P = 1$ . Everything else sits strictly between.

Example 1 — the base case · cell A

Worked example Ordinary count: one die

Statement: Roll one fair die. What is $P (result is a multiple of 3)$ ?

Forecast: Guess the fraction now — how many faces are multiples of $3$ , out of $6$ ?

Step 1 — Write the sample space. $S = {1, 2, 3, 4, 5, 6}, n (S) = 6.$ Why this step? Every calculation is a fraction of $S$ ; we cannot form the fraction until we know the denominator. Each face is equally likely because the die is fair — this is what lets us use $n (E) / n (S)$ .

Step 2 — Circle the event. Multiples of $3$ in $S$ : only $3$ and $6$ . $E = {3, 6}, n (E) = 2.$ Why this step? "Multiple of $3$ " is a rule; we apply it to each outcome and keep the ones that pass. Checking one at a time guarantees we miss none.

Step 3 — Divide. $P (E) = \frac{n ( E )}{n ( S )} = \frac{2}{6} = \frac{1}{3} .$

Verify: $\frac{1}{3} \approx 0.333$ , which sits inside $[0, 1]$ ✓. Sanity: exactly one of every three consecutive integers is a multiple of $3$ , so "about a third" is exactly right.

Example 2 — both degenerate cases at once · cells B and C

Worked example Impossible and sure events

Statement: Roll one fair die. Find (a) $P (face = 7)$ and (b) $P (face < 10)$ .

Forecast: One of these is never true and one is always true. Which extreme value does each get?

Step 1 — Same sample space as before. $S = {1, \dots, 6}$ , $n (S) = 6$ . Why this step? The experiment hasn't changed, only the events we ask about.

Step 2 (a) — Impossible event. No face equals $7$ , so the event is the empty set: $E_{a} = \emptyset, n (E_{a}) = 0, P (E_{a}) = \frac{0}{6} = 0.$ Why this step? An event that contains no outcomes is exactly the picture on the left of the figure — a blob covering nothing of the box. Probability $0$ means "cannot occur under these outcomes".

Step 3 (b) — Sure event. Every face is below $10$ , so the event is all of $S$ : $E_{b} = {1, 2, 3, 4, 5, 6} = S, n (E_{b}) = 6, P (E_{b}) = \frac{6}{6} = 1.$ Why this step? When the blob fills the entire box, $n (E) = n (S)$ and the fraction is $1$ . Probability $1$ means "certain". These two answers are the floor and ceiling of the whole subject.

Verify: $0$ and $1$ are exactly the endpoints of the allowed range $0 \leq P \leq 1$ ✓. Their sum with any other honest probability still stays in range.

Example 3 — the equally-likely trap · cell D

Worked example When you may NOT just count

Statement: Toss two fair coins. A friend argues: "the number of heads is $0$ , $1$ , or $2$ — three cases — so $P (exactly one head) = \frac{1}{3}$ ." Is the friend right? Find the true probability.

Forecast: Do you believe $\frac{1}{3}$ , or is a trap hiding?

Step 1 — Build the outcome sample space, not the summary one. First coin is H or T; for each, second coin is H or T: $S = {HH, HT, TH, TT}, n (S) = 4.$ Why this step? The formula $n (E) / n (S)$ is only legal when every listed outcome is equally likely. The four coin-pairs above are equally likely. But " $0$ , $1$ , or $2$ heads" are summaries that bundle different numbers of these — they are not equally likely, so counting them is illegal.

Step 2 — Circle the event on the legal space. "Exactly one head" = one H and one T = ${HT, TH}$ , so $n (E) = 2$ . Why this step? HT and TH are genuinely different outcomes (which coin showed the head), so both count.

Step 3 — Divide. $P (E) = \frac{2}{4} = \frac{1}{2} .$

Verify: List the head-counts with weights: $0$ heads = ${T T}$ (weight $1$ ), $1$ head = ${H T, T H}$ (weight $2$ ), $2$ heads = ${H H}$ (weight $1$ ). Weights $1 : 2 : 1$ sum to $4$ ✓, and the "one head" slice is $2/4 = \frac{1}{2}$ , not $\frac{1}{3}$ . The friend was wrong. See also Axioms of Probability — Axiom 2 forces the weights to sum to $1$ , which pins down each outcome's share.

Example 4 — "OR" and "AND" together · cells E and F

Worked example Union and intersection on one problem

Statement: Roll one fair die. Find (a) $P (even OR greater than 4)$ and (b) $P (even AND greater than 4)$ .

Forecast: "OR" gathers more outcomes, "AND" keeps fewer — which fraction is bigger?

Look at the overlap picture:

Step 1 — Name the two events. $A = even = {2, 4, 6}$ , $B = greater than 4 = {5, 6}$ , and $n (S) = 6$ . Why this step? "OR" and "AND" are operations between events, so we must have both events written out before combining them.

Step 2 (a) — Union for "OR". $A \cup B$ = every outcome in $A$ or $B$ , listed once: $A \cup B = {2, 4, 5, 6}, n = 4, P = \frac{4}{6} = \frac{2}{3} .$ Why this step? The number $6$ lives in both blobs (it is even and $> 4$ ). If we naively add $n (A) + n (B) = 3 + 2 = 5$ we count $6$ twice. The union rule keeps it once. In figure terms: the shaded "OR" region is the whole coloured area, and the overlap tile is not painted twice.

Step 3 (b) — Intersection for "AND". $A \cap B$ = outcomes in both: $A \cap B = {6}, n = 1, P = \frac{1}{6} .$ Why this step? "AND" demands both conditions, so only the overlap tile survives — the single face $6$ .

Verify: Inclusion–exclusion must hold: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ . Check: $\frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$ ✓. And "OR" ( $\frac{2}{3}$ ) is indeed bigger than "AND" ( $\frac{1}{6}$ ), as forecast. See Mutually Exclusive Events — if the overlap were empty, that subtracted term vanishes.

Example 5 — the complement shortcut · cell G

Worked example "At least one" via the opposite

Statement: Toss three fair coins. What is $P (at least one head)$ ?

Forecast: Would you rather list all the "at least one head" outcomes, or list the single opposite?

Step 1 — Size the sample space. Each coin: $2$ ways, three coins: $2 \times 2 \times 2$ (a counting-principle multiplication): $n (S) = 2^{3} = 8.$ Why this step? The denominator is fixed by how many independent binary choices we make.

Step 2 — Flip to the complement. "At least one head" is awkward (it covers $7$ of the $8$ outcomes). Its opposite complement is tiny: "no heads at all" = TTT. $E^{'} = {TTT}, n (E^{'}) = 1, P (E^{'}) = \frac{1}{8} .$ Why this step? Counting one outcome is far safer than counting seven. The complement rule lets us do the hard event through the easy one.

Step 3 — Subtract from $1$ . $P (at least one head) = 1 - P (E^{'}) = 1 - \frac{1}{8} = \frac{7}{8} .$ Why this step? An event and its complement together fill the whole box, so their probabilities add to $1$ ; rearranging gives $P (E) = 1 - P (E^{'})$ .

Verify: Direct count: outcomes with $\geq 1$ head are all except TTT, so $7$ of $8$ → $\frac{7}{8}$ ✓. Matches.

Example 6 — choosing ordered vs unordered $S$ · cell H

Worked example Two balls from a bag, without replacement

Statement: A bag holds three balls: Red, Blue, Green. Draw two, one after another, without replacement. What is $P (one of the two drawn is Red)$ ?

Forecast: Guess the fraction before choosing how to list outcomes.

Step 1 — Decide: ordered or unordered? Either works if used consistently. Take ordered (first ball, then second), because "without replacement" is naturally a sequence: $S = {RB, RG, BR, BG, GR, GB}, n (S) = 6.$ Why this step? Ordered listing makes "RB" and "BR" distinct, which is fine — as long as the event is counted the same ordered way, the ratio is unaffected. The danger is mixing an ordered $S$ with an unordered $E$ ; we avoid that by fixing the style up front.

Step 2 — Circle "Red appears". Outcomes containing R: RB, RG, BR, GR. $n (E) = 4.$ Why this step? R can be the first draw (RB, RG) or the second draw (BR, GR); an ordered space forces us to see both, so none is forgotten.

Step 3 — Divide. $P (E) = \frac{4}{6} = \frac{2}{3} .$

Verify (cross-check with unordered $S$ ). Unordered pairs: ${R, B}, {R, G}, {B, G}$ , so $n (S) = 3$ ; those containing R are ${R, B}, {R, G}$ , so $n (E) = 2$ ; ratio $= \frac{2}{3}$ ✓. Same answer because both numerator and denominator doubled — proof that either style is valid when kept consistent.

Example 7 — real-world word problem · cell I

Worked example Class committee

Statement: A class has $20$ students: $12$ girls and $8$ boys. One student is chosen at random to be class representative. What is the probability the representative is a boy?

Forecast: Boys are the smaller group — should the probability be under or over $\frac{1}{2}$ ?

Step 1 — Translate words into $S$ . The experiment is "pick one student"; every student is equally likely, so $n (S) = 20.$ Why this step? Real problems hide the sample space in a sentence. "Chosen at random" is the phrase that certifies equal likelihood — without it we could not use the counting formula.

Step 2 — Translate the question into $E$ . Favourable = "is a boy", and there are $8$ boys: $n (E) = 8.$ Why this step? The word "boy" is the membership rule; we simply count how many outcomes satisfy it.

Step 3 — Divide and simplify. $P (boy) = \frac{8}{20} = \frac{2}{5} = 0.4.$

Verify: $P (girl) = \frac{12}{20} = \frac{3}{5} = 0.6$ , and $0.4 + 0.6 = 1$ ✓ (a boy or a girl is certain). Also $0.4 < 0.5$ , matching the forecast that the smaller group is less likely.

Example 8 — exam twist: staged reasoning · cell J

Worked example Sum of two dice, then a condition

Statement: Roll two fair dice. Find (a) $P (sum = 8)$ , and (b) among all rolls, $P (sum = 8 AND both dice even)$ .

Forecast: Adding the extra "both even" demand can only shrink or hold the count — so which answer is smaller?

Step 1 — Fix the sample space of ordered pairs. First die $1$ – $6$ , second die $1$ – $6$ : $n (S) = 6 \times 6 = 36.$ Why this step? Two dice are two independent choices; the counting principle gives $36$ equally-likely ordered pairs $(first, second)$ .

Step 2 (a) — Count pairs summing to $8$ . Go through first die $= 1, \dots, 6$ and ask what the second must be: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2), n (E_{a}) = 5.$ Why this step? Listing by the first die guarantees completeness and no repeats. Note $(1, 7)$ is impossible (no $7$ ), which is why we get $5$ and not $6$ . $P (E_{a}) = \frac{5}{36} .$

Step 3 (b) — Add the second condition. From those five pairs, keep only the ones where both entries are even: $(2, 6), (4, 4), (6, 2), n (E_{b}) = 3.$ Why this step? "AND both even" is an intersection: an outcome must pass two filters — sum $= 8$ and even–even. We reuse the list from Step 2 instead of restarting, which is faster and error-proof. $P (E_{b}) = \frac{3}{36} = \frac{1}{12} .$

Verify: $\frac{3}{36} < \frac{5}{36}$ — the extra condition shrank the count, exactly as forecast ✓. Sanity on the survivors: $2 + 6 = 8$ , $4 + 4 = 8$ , $6 + 2 = 8$ , all even entries ✓.

Recall check

Recall Which rule for which word?

"Exactly one head from two coins" — legal to count directly? ::: Yes, on $S = {H H, H T, T H, T T}$ these four are equally likely; answer $\frac{1}{2}$ . Why is $P (sum = 7)$ built on $36$ outcomes, not $11$ ? ::: The $11$ possible sums are not equally likely; the $36$ ordered pairs are. Fast way to do "at least one" ::: Use the complement: $1 - P (none)$ . " $A$ OR $B$ " count fix when they overlap ::: Take the union and count each shared outcome only once (inclusion–exclusion).

Prerequisite / next links: Parent topic · Sample Space · Events and Set Theory · Complement of an Event · Mutually Exclusive Events · Axioms of Probability · Counting Principles · Conditional Probability · Random Variables · Law of Large Numbers