Probability with Coins, Dice, and Cards
Understanding Chance and Likelihood
Essential Understanding: Probability measures how likely an event is to occur. It is expressed as a number between 0 (impossible) and 1 (certain). Coins, dice, and cards are classic tools for learning probability because their outcomes are well-defined and equally likely.
The Probability Formula
Basic Probability
The probability of an event \(A\) occurring is:
Probability is always between 0 and 1: \( 0 \leq P(A) \leq 1 \)
The Probability Scale
Impossible 0.25
Unlikely 0.5
Even chance 0.75
Likely 1
Certain
Sample Space
The sample space (denoted \(S\)) is the set of all possible outcomes of an experiment.
It tells us everything that could possibly happen when we perform the experiment.
Probability with Coins
Fair Coin
A fair coin has 2 equally likely outcomes:
| Event | Favourable Outcomes | Probability |
|---|---|---|
| Getting Heads | 1 (H) | \( P(H) = \frac{1}{2} = 0.5 \) |
| Getting Tails | 1 (T) | \( P(T) = \frac{1}{2} = 0.5 \) |
Coin Flip Simulator
Flip a virtual coin and track the results! Watch how the experimental probability approaches \(\frac{1}{2}\) with more flips.
Probability with Dice
Fair Six-Sided Die
A fair die has 6 equally likely outcomes:
| Event | Favourable Outcomes | Probability |
|---|---|---|
| Rolling a 4 | 1 outcome: {4} | \( P(4) = \frac{1}{6} \) |
| Rolling an even number | 3 outcomes: {2, 4, 6} | \( P(\text{even}) = \frac{3}{6} = \frac{1}{2} \) |
| Rolling a number > 4 | 2 outcomes: {5, 6} | \( P(>4) = \frac{2}{6} = \frac{1}{3} \) |
| Rolling a prime number | 3 outcomes: {2, 3, 5} | \( P(\text{prime}) = \frac{3}{6} = \frac{1}{2} \) |
Dice Roll Simulator
Roll a virtual die and track the distribution of outcomes!
Total rolls: 0
Probability with Playing Cards
Standard Deck of 52 Cards
A standard deck contains 52 cards divided into:
Red Suits (26 cards)
♥ Hearts (13 cards)
♦ Diamonds (13 cards)
Black Suits (26 cards)
♠ Spades (13 cards)
♣ Clubs (13 cards)
Each suit contains 13 cards: A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K
| Event | Favourable | Total | Probability |
|---|---|---|---|
| Drawing a Heart | 13 | 52 | \( \frac{13}{52} = \frac{1}{4} \) |
| Drawing a King | 4 | 52 | \( \frac{4}{52} = \frac{1}{13} \) |
| Drawing a red card | 26 | 52 | \( \frac{26}{52} = \frac{1}{2} \) |
| Drawing a face card (J, Q, K) | 12 | 52 | \( \frac{12}{52} = \frac{3}{13} \) |
| Drawing the Ace of Spades | 1 | 52 | \( \frac{1}{52} \) |
Worked Examples
Question: Two fair dice are thrown. Find the probability that the sum of the numbers is 7.
Each die has 6 outcomes, so total = \( 6 \times 6 = 36 \) outcomes
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 outcomes
\( P(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6} \)
Question: A card is drawn at random from a standard deck. Find the probability of drawing a red King.
Total cards in deck = 52
Red Kings = King of Hearts + King of Diamonds = 2 cards
\( P(\text{red King}) = \frac{2}{52} = \frac{1}{26} \)
Question: Two fair coins are tossed. Find the probability of getting at least one Head.
\( S = \{HH, HT, TH, TT\} \) = 4 outcomes
{HH, HT, TH} = 3 outcomes
\( P(\text{at least one H}) = \frac{3}{4} \)
Alternative method: \( P(\text{at least one H}) = 1 - P(\text{no Heads}) = 1 - \frac{1}{4} = \frac{3}{4} \)
CSEC Practice Questions
Test Your Understanding
\( P(<3) = \frac{2}{6} = \frac{1}{3} \)
\( P(\text{Q or K}) = \frac{8}{52} = \frac{2}{13} \)
Exactly 2 Heads: {HHT, HTH, THH} = 3 outcomes
\( P(\text{exactly 2H}) = \frac{3}{8} \)
Total outcomes = 36
\( P(\text{doubles}) = \frac{6}{36} = \frac{1}{6} \)
Not red = Blue + Green = 3 + 2 = 5
\( P(\text{not red}) = \frac{5}{10} = \frac{1}{2} \)
Or use: \( P(\text{not red}) = 1 - P(\text{red}) = 1 - \frac{5}{10} = \frac{1}{2} \)
Key Points to Remember
- \( P(A) = \frac{\text{Favourable outcomes}}{\text{Total outcomes}} \)
- Probability is always between 0 and 1
- Coin: 2 outcomes (H, T)
- Die: 6 outcomes (1, 2, 3, 4, 5, 6)
- Cards: 52 cards, 4 suits × 13 values
- \( P(\text{not } A) = 1 - P(A) \) — The complement rule
- For multiple coins/dice, multiply the number of outcomes: 2 coins = \(2^2 = 4\), 2 dice = \(6^2 = 36\)
- Always simplify fractions — Write \(\frac{1}{4}\) not \(\frac{13}{52}\)
- Read carefully — "at least one" is different from "exactly one"
- List the sample space when working with small experiments
- Use the complement when it's easier: P(at least one) = 1 - P(none)
- For cards: Remember there are 4 of each value (4 Kings, 4 Aces, etc.)
- Check your answer: Probability cannot be greater than 1 or negative