Probability

Definition

P(E) = favourable outcomes / total outcomes, always between 0 and 1. P(not E) = 1 - P(E). Target: 45 seconds.

Sample spaces to remember

Sums with two dice

Rules

• Addition rule: P(A or B) = P(A) + P(B) - P(A and B).
• Independent events: P(A and B) = P(A) x P(B).
• At least one: 1 - P(none). Problem solved by at least one of A, B, C = 1 - (1 - a)(1 - b)(1 - c).
• Without replacement: use combinations (nCr) or reduce the count after each draw.

Worked example

Two dice thrown: P(sum = 7) = 6/36 = 1/6.

Worked example 2

A card is drawn: P(face card) = 12/52 = 3/13.

Worked example 3

A bag has 3 red and 5 black balls; 2 are drawn. P(both red) = 3C2/8C2 = 3/28.

Worked example 4

Three coins: P(at least one head) = 1 - 1/8 = 7/8.

Common traps

• Double counting overlap (king or heart = 16 cards, not 17).
• Assuming replacement when the question draws together.
• Adding probabilities of events that can happen together.
• Forgetting that (2, 6) and (6, 2) are different outcomes for two dice.

More examples

• Leap year with 53 Sundays: 366 days = 52 weeks + 2 days. The 2 extra days can be any of 7 consecutive pairs, 2 of which contain a Sunday. P = 2/7.
• Two aces without replacement: 4C2/52C2 = 6/1,326 = 1/221.
• Numbers 1 to 20 divisible by 3 or 5: 6 multiples of 3 + 4 multiples of 5 - 1 (the number 15) = 9, so P = 9/20.
• Odds: odds in favour of an event a : b means P = a/(a + b); odds against a : b means P = b/(a + b).

Choosing the method

1. Write the total number of outcomes first (36 for two dice, nCr for draws).
2. Count favourable cases directly if they are few; otherwise count the complement.
3. For independent events in sequence, multiply the probabilities of each stage.