Classical probability
When all outcomes are equally likely, P(E) is n(E)/n(S): favourable outcomes over total outcomes. Every probability lies between 0 and 1.
Classical probability and its rules
- If all outcomes of a random experiment are equally likely, the probability of an event E is P(E) = n(E)/n(S).
- n(E) is the number of outcomes favourable to E; n(S) is the total number of outcomes.
- 0 ≤ P(E) ≤ 1 for every event.
- P(S) = 1 for the sure event, and P(∅) = 0 for the impossible event.
- P(E) + P(Ē) = 1, so P(Ē) = 1 − P(E).
- Simplify the fraction. You may also write it as a decimal.
Two dice: all 36 sums (first die down, second die across)
| + | 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 |
Example 1: two dice
Two dice are rolled. Find the probability that (i) the sum is 8 (ii) both dice show the same number.
- n(S) = 6 × 6 = 36.
- (i) Sum 8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), so n(A) = 5.
- P(A) = 5/36
- (ii) Doublets: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), so n(B) = 6.
- P(B) = 6/36 = 1/6
(i) 5/36 (ii) 1/6
Example 2: a pack of cards
A card is drawn from a well-shuffled pack of 52 cards. Find the probability that it is (i) a red king (ii) a face card (iii) not a spade.
- n(S) = 52.
- (i) Red kings: the king of hearts and the king of diamonds, so 2. P = 2/52 = 1/26
- (ii) Face cards: J, Q and K in 4 suits = 12. P = 12/52 = 3/13
- (iii) There are 13 spades, so P(spade) = 13/52 = 1/4.
- P(not a spade) = 1 − 1/4 = 3/4
(i) 1/26 (ii) 3/13 (iii) 3/4
Example 3: 53 Fridays in a leap year
Find the probability that a leap year chosen at random has 53 Fridays.
- A leap year has 366 days = 52 weeks + 2 days.
- The 52 weeks give 52 Fridays. The 2 extra days decide the 53rd.
- The 2 extra days can be (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat) or (Sat, Sun). So n(S) = 7.
- A Friday comes in (Thu, Fri) and (Fri, Sat), so n(E) = 2.
- P(53 Fridays) = 2/7
2/7
The coin toss before a cricket match is fair because each captain has a probability of 1/2 of winning it. In Ludo, the chance of rolling a 6 to start is 1/6.
1 marks: “What is the probability of an impossible event?” 0.
2 marks: “A bag has 5 red, 4 blue and 3 green balls. One ball is taken at random. Find the probability that it is not blue.” n(S) = 12, P(blue) = 4/12 = 1/3. P(not blue) = 1 − 1/3 = 2/3.
5 marks: “Two dice are rolled together. Find the probability that (i) the sum is greater than 9 (ii) the sum is a prime number (iii) the product of the numbers is 12.” n(S) = 36. (i) Sums 10, 11, 12: 6 outcomes, P = 1/6. (ii) Prime sums 2, 3, 5, 7, 11: 15 outcomes, P = 5/12. (iii) (2, 6), (3, 4), (4, 3), (6, 2): P = 4/36 = 1/9.
Probability = what I want over all that can happen: n(E)/n(S). It never goes below 0 or above 1.