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Learn · Mathematics · Grade 10 · Probability

Probability: sample spaces and events

Probability as favourable over total, sample spaces for coins, dice and cards, complementary events, "or" with the overlap removed, and independent "and".

High school Lesson 15 of 31 Interactive figure Practice

First, the rules

Favourable over total

When every outcome is equally likely, probability is a fraction: how many outcomes you want, over how many there are. Everything else is bookkeeping.

★ Sample space and events
Examples 1, 2
The sample space S lists every possible outcome; an event E is a subset of S. P(E) = s(E) / s(S). Always 0 ≤ P(E) ≤ 1.
★ Complement
Example 3
P(E′) = 1 − P(E)
"At least one" is often easiest as 1 − P(none).
Addition rule ("or")
Example 2
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Mutually exclusive events have P(A ∩ B) = 0, so you just add.
★ Independent events ("and")
Example 3
P(A ∩ B) = P(A) · P(B)
Two dice, two coins, drawing WITH replacement.
Counting feeds probability
Example 3
P = C(favourable)C(total)
Choosing 2 red marbles from a bag: C(red, 2) / C(all, 2).
Two dice
Example 2
36 equally likely pairs (first die, second die). The sum 7 appears in 6 of them; the sums 2 and 12 in one each.

See it

The 36 outcomes of two dice

Slide the target sum. The highlighted cells are the favourable outcomes; the probability is their number over 36. Notice how 7 is the most likely sum and 2 or 12 the least.

Step by step

Worked examples

1

One die, one card

favourable / total

(a) A fair die is rolled. Find the probability of a prime number. (b) A card is drawn from a deck of 52. Find the probability of a face card (J, Q or K).

Ask yourself: List the favourable outcomes explicitly before you count.

Show the solution
  1. (a) Primes on a die: {2, 3, 5}. P = 36 = 12.
  2. (b) Face cards: 3 per suit × 4 suits = 12. P = 1252 = 313.

Answer 12 and 313

2

Two dice: "or"

addition rule

Two dice are rolled. Find the probability that the sum is 7 or at least one die shows a 6.

Ask yourself: The two events overlap (a 1 and a 6 gives sum 7 AND a six). Count the overlap so you do not double-count.

Show the solution
  1. Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — 6 outcomes, P(A) = 636.
  2. At least one 6: 11 outcomes (the 6th row and 6th column of the table minus the shared (6,6)), P(B) = 1136.
  3. Both: (1,6) and (6,1), P(A ∩ B) = 236.
  4. P(A ∪ B) = 6 + 11 − 236 = 1536 = 512

Answer 512

3

Marbles and "at least one"

combinations · complement

A bag holds 4 red and 6 blue marbles. Two are drawn at once. Find (a) the probability both are red, (b) the probability at least one is red.

Ask yourself: Drawing two at once is a combination. For "at least one", what is the opposite event?

Show the solution
  1. Total pairs: C(10, 2) = 45.
  2. (a) Both red: C(4, 2) = 6, so P = 645 = 215.
  3. (b) Opposite of "at least one red" is "both blue": C(6, 2) = 15, P(both blue) = 1545 = 13.
  4. P(at least one red) = 1 − 13 = 23

Answer 215 and 23

Practice: probability

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Dice, cards, marbles, coins and complements — pick the exact fraction. Your best score is kept in this browser only.

About this lesson

Probability basics: sample space and events, P(E) = favourable / total for equally likely outcomes, the complement rule P(E′) = 1 − P(E), the addition rule P(A ∪ B) = P(A) + P(B) − P(A ∩ B), independent events and the multiplication rule, two dice tables, and how counting (combinations) feeds into probability. Worked examples and practice with exact fractions.

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