Lesson 16: Probability Basics

Will it rain tomorrow? Will you roll a six? Probability is the mathematics of "how likely" — it turns hunches into numbers between 0 and 1. A probability of 0 means impossible, 1 means certain, and everything real lives in between: a coin flip at 1/2, a die roll at 1/6, a forecast at 70%. In this lesson you will compute probabilities for coins, dice, spinners and bags of marbles, meet theoretical versus experimental probability, and use the complement trick for "not" and "at least" questions.

Theoretical probability

When all outcomes are equally likely, probability = (number of favourable outcomes) ÷ (total number of possible outcomes). The result can be written as a fraction, decimal or percent — the same three languages as lesson 4.

Worked example 1. Rolling a fair six-sided die: P(rolling a 4) = 1/6. P(even number) = 3/6 = 1/2 (the even faces are 2, 4, 6). P(rolling more than 4) = 2/6 = 1/3 (only 5 and 6 qualify).

Worked example 2. A bag holds 4 red, 3 blue and 5 green counters — 12 in total. P(red) = 4/12 = 1/3. P(green) = 5/12. Notice all three probabilities sum to 1: 4/12 + 3/12 + 5/12 = 12/12 = 1. That is always true for a complete list of outcomes, and it is a powerful self-check.

The complement: "not" and "at least one"

Sometimes it is easier to compute the opposite. The complement rule says P(not A) = 1 − P(A). For the bag above, P(not blue) = 1 − 3/12 = 3/4 — quicker than counting red plus green, and identical (9/12 = 3/4 ✓). Weather apps use the same idea: if P(rain) = 0.3, then P(no rain) = 1 − 0.3 = 0.7.

Experimental probability

Theoretical probability predicts what should happen; experimental probability records what did happen: (times the event occurred) ÷ (times we tried). A coin flipped 50 times landing on heads 28 times gives an experimental probability of 28/50 = 0.56 — close to, but not exactly, the theoretical 0.5. The more trials you run, the closer experimental results usually settle to the theoretical value (the law of large numbers). This is why casinos are calm: over millions of games, the maths always shows up.

In the real world: games, weather and fairness

Game designers tune probabilities to make games feel fair but exciting — a 1/6 chance of a bonus roll feels rare and rewarding; 1/2 would feel cheap. Board-game spinners numbered 1 to 8 give P(odd) = 4/8 = 1/2. Insurers, doctors and forecasters all speak probability daily: "a 30% chance of rain" means rain fell on about 3 of the last 10 days with similar conditions. Understanding the scale from 0 to 1 keeps you from being fooled — a 1-in-1,000 risk and a 1-in-10 risk sound similar in a headline and are wildly different in life.

Common mistakes

• Forgetting to simplify. P(red) = 4/12 is correct but unfinished — give 1/3. Always cancel to lowest terms.
• Counting outcomes that are not equally likely. "Rolling a total of 2 or 12 with two dice" is not 2/12 — the 36 equally likely pairs make each of those totals just 1/36. Check that your outcomes really are equally likely before counting.
• Probabilities bigger than 1. If your calculation gives 7/5, something is wrong — no probability exceeds 1.
• Treating experimental results as exact predictions. 28 heads in 50 flips does not mean the coin is biased toward 0.56; small samples wobble. More trials give a steadier estimate.

Exam Experience

1. A bag contains 2 red marbles and 6 blue marbles. A marble is chosen at random. What is P(red)? A) 1/3 B) 1/4 C) 1/6 D) 3/4

2. A fair die is rolled. What is the probability of rolling a prime number (2, 3 or 5)? A) 1/6 B) 1/3 C) 1/2 D) 5/6

3. The probability that a bus is late is 0.15. What is the probability that it is not late? A) 0.15 B) 0.75 C) 0.85 D) 1.15

4. (Constructed response) A gumball machine contains 6 cherry, 4 lemon and 10 grape gumballs. A gumball is dispensed at random. (a) Find the probability that it is cherry, giving your answer as a simplified fraction. (b) Find the probability that it is lemon or grape. (c) If 50 gumballs are dispensed, use probability to predict how many will be lemon. Show your reasoning.

Answer key and explanations

• 1. B — There are 8 marbles in total, so P(red) = 2/8 = 1/4.
• 2. C — Three of the six faces (2, 3, 5) are prime: 3/6 = 1/2.
• 3. C — Complement: 1 − 0.15 = 0.85.
• 4. Total = 6 + 4 + 10 = 20. (a) P(cherry) = 6/20 = 3/10. (b) P(lemon or grape) = (4 + 10)/20 = 14/20 = 7/10. (c) P(lemon) = 4/20 = 1/5, so in 50 gumballs expect about 1/5 × 50 = 10 lemon.

Scoring guide for question 4 (3 points)

• 1 point for the total (20) and P(cherry) = 3/10, simplified.
• 1 point for P(lemon or grape) = 7/10 (adding the favourable counts before dividing, or adding 4/20 + 10/20).
• 1 point for the prediction of 10 lemon gumballs, from P(lemon) = 1/5 applied to 50 trials, with reasoning shown.

Going further: two coins and honest experiments

Worked example 3 (two coins). Flip two coins. The equally likely outcomes are HH, HT, TH, TT — four of them. P(two heads) = 1/4; P(exactly one head) = 2/4 = 1/2; P(at least one head) = 3/4, found instantly by the complement: 1 − P(no heads) = 1 − 1/4 = 3/4. Listing outcomes in a tidy grid is the professional way to make "equally likely" visible.

Worked example 4. With one die, P(rolling 1 or 6) = 2/6 = 1/3, and P(not rolling a 6) = 1 − 1/6 = 5/6. "Or" between separate outcomes adds their counts; "not" subtracts from 1. Those two moves handle most probability questions at this level.

Worked example 5. A standard deck of cards has 52 cards, 26 red and 26 black. P(red card) = 26/52 = 1/2. P(heart) = 13/52 = 1/4. Card probabilities are just favourable ÷ total, with a bigger deck of outcomes to count carefully.

Worked example 6 (is the spinner fair?). A spinner with 4 equal sections (theoretical P(red) = 1/4) is spun 200 times and lands on red 45 times. Experimental probability = 45/200 = 0.225, close to 0.25. With 200 trials, that gap is ordinary wobble — no evidence of unfairness. Had red appeared 90 times (0.45), suspicion would be justified. Judging how far is too far starts with comparing the two kinds of probability exactly like this.

Quick practice

• 1. A bag has 7 green and 3 yellow counters. Find P(yellow).
• 2. Write P(heads on a fair coin) as a percent.
• 3. What number means "certain"? What number means "impossible"?
• 4. A spinner numbered 1 to 4: find P(odd).

Answers. 1. 3/10 (3 yellow out of 10 counters). 2. 50% (1/2). 3. Certain = 1; impossible = 0 — the two ends of the probability scale. 4. Odd numbers are 1 and 3: 2/4 = 1/2.

Investigation: designing for a probability

A fairground stall wants a spinner where P(red) is exactly 3/8. Design it: divide the spinner into 8 equal sections and colour 3 of them red. Equal sections are the whole secret — probability targets are met by counting equal outcomes, so the designer's job is choosing the denominator (8) and the favourable count (3). Make P(red) = 5/12 instead and you need 12 sections, 5 red. Now roll two dice and hunt the most famous total: 7. The 36 equally likely pairs include (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — six ways — so P(7) = 6/36 = 1/6, the most likely total of all (6, 7 and 8 dominate dice games). Finally, plan an experiment: to test whether a coin is fair, flip it 60 times and expect about 30 heads. Getting 27 or 34 is wobble; getting 48 would be evidence. Designing the game, predicting the result, and judging the outcome — that is the complete probability cycle, and you have now walked around all of it.

Challenge corner

From a standard deck of 52 cards, P(drawing an ace) = 4/52 = 1/13 — four aces among 52 cards, simplified by dividing by 4. It is worth pausing on how small that is: in 13 random draws you would expect roughly one ace, which is why aces feel special in card games. Second: a spinner divided into 12 equal sections numbered 1 to 12. The multiples of 3 are 3, 6, 9, 12 — four of them — so P(multiple of 3) = 4/12 = 1/3. Notice that "numbered 1 to 12" quietly means 12 equally likely outcomes; whenever a problem hands you a numbered or sectioned device, your first move is counting its equally likely outcomes, and your second is counting the favourable ones. Probability problems are counting problems in disguise — the fraction writes itself once the counting is honest.