List everything that could happen. Count the ones you want. That is the whole method.
Find the probability
Two ordinary dice are rolled and the numbers are added. What is the probability the total is 5?
There are 6 × 6 = 36 equally likely outcomes. A total of 5 happens four ways — (1,4), (2,3), (3,2), (4,1) — so the probability is 436 = 19. A treats the eleven possible totals 2 to 12 as equally likely, which they are not: there is only one way to make 2, but four ways to make 5. D counts five outcomes instead of four by including (0,5), which no die can produce.
| Word | Means | Rolling one die |
|---|---|---|
| Experiment | the thing you do | rolling it once |
| Outcome | one possible result | a 4 |
| Sample space | every possible outcome | 1, 2, 3, 4, 5, 6 |
| Event | the outcomes you care about | "an even number" = 2, 4, 6 |
An event is usually several outcomes bundled together by a description. "Rolling more than 4" is one event made of two outcomes.
P(event) = favourable outcomes / total outcomes
P(even on one die)
the event
= 36
2, 4, 6 out of six faces
= 12
simplify
Every answer lands between 0 and 1. An answer of 65, or a negative one, means the favourable count went above the total — go back and recount.
Every cell is one outcome, and every cell is equally likely.
| + | 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, four of them showing 5. So P(total of 5) = 436 = 19 — the opening problem, answered by counting.
The table also explains why 7 is the commonest total: six cells show it, more than any other number. Board games have known this for centuries.
P(exactly one head) = 24 = 12, because HT and TH both qualify. HT and TH are different outcomes — that is the step most often missed.
Each extra toss splits every branch in two again: three tosses give 2 × 2 × 2 = 8 outcomes, from HHH to TTT, and four tosses give 16.
P(not A) = 1 − P(A)
A jar holds 5 green, 4 white and 3 yellow candies. One is picked at random. What is the probability it is not white?
P(white) = 412 = 13
12 candies in all
P(not white) = 1 − 13 = 23
everything else
Check by counting directly: 5 green + 3 yellow = 8, and 812 = 23 ✓
Two dice are rolled. What is the probability that at least one of them shows a 6?
no 6 on either: 5 × 5 = 25 outcomes
each die has 5 faces that are not 6
P(at least one 6) = 1 − 2536 = 1136
the rest of the 36 cells
In the table of step 3, the 6-row and the 6-column hold 6 + 6 − 1 = 11 cells; the corner (6, 6) belongs to both.
16 + 16 = 1236 counts the double six twice — once for each die. The true count is 11 cells, so the answer is 1136, not 13.
Still favourable over total; these are the answers at and beyond the ends.
One die is rolled. Find P(7) and P(a number less than 7).
P(7) = 06 = 0
no face shows 7: impossible
P(less than 7) = 66 = 1
every face qualifies: certain
0 and 1 are the two ends of the probability line, and both can really happen as answers.
A spinner has 10 equal sections numbered 1 to 10. What is the probability of a prime?
primes: 2, 3, 5, 7
1 is not prime, and 9 = 3 × 3 is not either
P(prime) = 410 = 25
simplify
Listing the event carefully is half the work; counting 1 or 9 by habit gives 510 or 610.
A fair die is rolled 300 times. About how many sixes should we expect?
300 × 16 = 50
probability × number of trials
about 50, not exactly 50
the experimental count will wander a little
The expected count is a forecast; 46 or 55 sixes would be perfectly normal.
Experimental probability = times it happened / times you tried
More trials pull the experimental value towards the theoretical one. A coin that came up heads five times running is not "due" a tail — it has no memory of the last five tosses.
Probability items use a die, coins, a spinner or a bag of marbles and ask for one event, a "not" or "at least one", or how an experiment compares with theory. The fastest route is to list or tabulate the sample space and count; the 10-second check is that the answer sits between 0 and 1 and that P(event) + P(not event) = 1.
Find the probability
A coin is tossed three times. What is the probability of getting exactly two heads?
Three tosses give 2 × 2 × 2 = 8 outcomes. Exactly two heads happens three ways — HHT, HTH and THH — so the probability is 38. A counts only HHT and forgets that the tail can land in any of the three positions. B uses 28 and misses one arrangement.
Hint: write all eight outcomes out in a list before counting. "Exactly two" does not include HHH.
