1
The sample space is everything that could happen.
| Word | Means | In our problem |
|---|
| outcome | one single result | (3, 4) — a three then a four |
|---|
| sample space | the list of all outcomes | all 36 pairs |
|---|
| event | the outcomes you care about | the ones summing to 7 |
|---|
P(event) = outcomes in the eventoutcomes in the sample space
Only if the outcomes are equally likely
That fraction assumes every item on the list has the same chance. It is why the
list must be of individual results — a six then a one — and not of summaries
like “a sum of seven”.
2
Count it before you list it.
The counting principle tells you how long the list should be, so you know when you have missed one.
6 × 6 = 36 outcomes
Two coins give 2 × 2 = 4. Three coins give 8. A coin and a die give 12. Every
extra stage multiplies, so a sample space grows faster than you expect — which
is exactly why a system beats writing them down as they occur to you.
3
Two dice: use the grid.
Rows for the first die, columns for the second, sums in the middle.
| + |
1 | 2 | 3 |
4 | 5 | 6 |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
|---|
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|
36 cells, one per outcome — and the sevens fall on a clean diagonal.
(1,6) (2,5) (3,4) (4,3) (5,2) (6,1)
six ways
P = 636 = 16
divide both by 6
4
(3, 4) and (4, 3) are two outcomes.
Even when the dice are the same colour.
sums to 7
6 cells
P = 16
sums to 2
1 cell — only (1,1)
P = 136
sums to 11
(5,6) and (6,5)
P = 236 = 118
Eleven sums are not eleven equal chances
The possible totals run 2 to 12, so it is tempting to say P(7) = 111. But a 7
happens six ways and a 2 only one — the totals are not equally likely, and the
36 individual rolls are.
5
Variations: “at least one”, four coins, no replacement.
At least one six. Two dice are rolled. What is the probability that at least one shows a 6?
first die 6: (6,1) … (6,6) — 6 cells
the bottom row of the grid
second die 6, first not: 5 more cells
(6,6) is already counted
P = 1136
the row and column cross once
What is different: the event is a row plus a column of the grid, and they share one cell.
Answering 13
16 + 16 = 1236 counts (6, 6) twice. Count the cells, not the dice.
Four coins. Four coins are tossed. What is the probability of at least 3 heads?
2 × 2 × 2 × 2 = 16 outcomes
too many to guess at
exactly 3: HHHT, HHTH, HTHH, THHH — 4
the T can sit in 4 places
exactly 4: HHHH — 1
“at least 3” includes 4
P = 4 + 116 = 516
add the cases
What is different: list only the event, by where the odd one out sits, instead of all sixteen outcomes.
No replacement. Two numbers are drawn one after the other, without replacement, from {1, 2, 3, 4, 5}. What is the probability that their sum is even?
5 × 4 = 20 ordered outcomes
no (3, 3): a number cannot come twice
both odd: 3 × 2 = 6; both even: 2 × 1 = 2
even sums need matching parity
P = 820 = 25
lower than the 1325 with replacement
What is different: the grid loses its diagonal, so the second draw has one choice fewer.
6
Coins: list them in a fixed order.
Three coins, 2 × 2 × 2 = 8 outcomes. Write all the H-first ones, then all the T-first ones.
| Outcome | Heads | Outcome | Heads |
|---|
| HHH | 3 | THH | 2 |
|---|
| HHT | 2 | THT | 1 |
|---|
| HTH | 2 | TTH | 1 |
|---|
| HTT | 1 | TTT | 0 |
|---|
exactly 2 heads: HHT, HTH, THH
three of them
The alphabet trick
Listing HHH, HHT, HTH, HTT, THH, THT, TTH, TTT is just counting in binary with
H for 0 and T for 1. Follow that order and skipping one becomes almost
impossible.
On the exam
Sample-space items are usually two dice, a coin and a die, or three coins, with an event such as “the sum is prime”, “the product is odd” or “the numbers differ by 2”. Multiply to get the size, list the event in a fixed order, and remember that (2, 5) and (5, 2) are two outcomes. Shortcut: a product is odd only when every number is odd, so count the odd × odd outcomes and subtract them from the whole when the question asks for “even”.
✓
Check: the whole space adds to 1.
Count the cells for every total. If they do not reach 36, something was missed.
2 to 6
1 + 2 + 3 + 4 + 5 = 15
below seven
7
6
the diagonal
8 to 12
5 + 4 + 3 + 2 + 1 = 15
15 + 6 + 15 = 36 ✓
Remember the flow
Multiply for the size · List in a system · Count the wins
Keep the order — (3,4) is not (4,3).
