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If you flip a fair coin, what is the chance it lands heads? How would you write that as a number between 0 and 1? What about the chance of rolling a 6 on a standard die?
Probability gives us a precise way to measure how likely something is to happen. From flipping coins to predicting weather, probability is the mathematics of uncertainty — and it's everywhere.
Probability is always between 0 and 1 inclusive. A probability greater than 1 or less than 0 is impossible — you have made an arithmetic error. Also: "50-50" does NOT mean every situation with two options is equally likely. Two options can be very different in likelihood (e.g. you win the lottery or you don't — but they're not 50-50!).
For experiments where all outcomes are equally likely:
$$P(\text{event}) = \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$$This fraction is always between 0 (impossible) and 1 (certain).
Examples:
The sample space is the set of all possible outcomes. Always list it before calculating probability.
For combined experiments (e.g. coin AND die), the total number of outcomes = number of outcomes for each experiment multiplied together. Coin and die: $2 \times 6 = 12$ outcomes.
Listing the full sample space guarantees you count favourables correctly — it's worth doing every time.
1. Rolling a fair die:
$P(\text{even}) = \dfrac{3}{6} = \dfrac{1}{2}$ (favourable: {2, 4, 6}, total: {1,2,3,4,5,6})
$P(\text{greater than 4}) = \dfrac{2}{6} = \dfrac{1}{3}$ (favourable: {5, 6})
2. Cards from a standard 52-card deck:
$P(\text{red card}) = \dfrac{26}{52} = \dfrac{1}{2}$ (26 red cards)
$P(\text{King}) = \dfrac{4}{52} = \dfrac{1}{13}$ (4 Kings in the deck)
3. Random letter from A to Z:
$P(\text{vowel}) = \dfrac{5}{26}$ (vowels: A, E, I, O, U)
The complement of an event $A$ is everything that is NOT $A$. It is written $A'$ or "not A".
$$P(\text{not } A) = 1 - P(A)$$This works because the event and its complement cover all possibilities — exactly one of them must happen. So their probabilities sum to 1.
Examples:
Probability — key formulas:
$$P(\text{event}) = \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}} \qquad 0 \le P \le 1$$
$$P(\text{not } A) = 1 - P(A) \qquad \text{(complementary events)}$$
Expected outcomes in $n$ trials $= n \times P(\text{event})$
What is the probability of rolling an odd number on a standard fair die?
A bag contains 3 red, 4 blue, and 3 green marbles. What is the probability of drawing a blue marble?
The probability of winning a game is 0.35. What is the probability of NOT winning?
How many outcomes are in the sample space when two standard dice are rolled together?
A bag has 4 red and 6 blue counters. What is the probability of NOT drawing a red counter?
Q6. A bag contains 5 red, 3 blue, and 2 green marbles.
(a) Find P(red).
(b) Find P(blue).
(c) Find P(not green).
(d) Add your answers to (a), (b), and the probability of drawing green together. What do you notice?
Q7. A fair coin is flipped and a standard die is rolled at the same time.
(a) List the complete sample space (all 12 outcomes) using the format H1, H2, T1, T2, etc.
(b) Find P(head AND even number).
(c) Find P(tail AND a number greater than 4).
Q8. The probability of winning a prize at a school raffle is 0.15.
(a) Find P(not winning a prize).
(b) If 200 students each buy one ticket, estimate how many students win a prize.
(c) If 200 students each buy one ticket, estimate how many students do NOT win a prize.
Total marbles = 5 + 3 + 2 = 10.
(a) P(red) = 5/10 = 1/2.
(b) P(blue) = 3/10.
(c) P(green) = 2/10 = 1/5; P(not green) = 1 − 1/5 = 4/5 = 8/10.
(d) P(red) + P(blue) + P(green) = 5/10 + 3/10 + 2/10 = 10/10 = 1. All probabilities sum to 1 — every marble must be one colour.
(a) Sample space: {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6} — 12 outcomes.
(b) Head AND even: {H2, H4, H6} — 3 outcomes. P = 3/12 = 1/4.
(c) Tail AND greater than 4: {T5, T6} — 2 outcomes. P = 2/12 = 1/6.
(a) P(not winning) = 1 − 0.15 = 0.85.
(b) Expected winners = 200 × 0.15 = 30 students.
(c) Expected non-winners = 200 × 0.85 = 170 students.
A bag has $x$ red marbles and 6 blue marbles.
(a) Write an expression for the total number of marbles in the bag.
(b) Given that $P(\text{red}) = \dfrac{2}{5}$, set up and solve an equation to find the value of $x$.
(c) Find P(not red) — use your value of $x$.
(d) If 3 more blue marbles are added to the bag, what is the new P(red)? Show your working.
(e) Is the new P(red) higher or lower than the original? Why does this make sense?