Simple event probability calculations

Every problem in this topic — regardless of whether it involves a coin, a die, a card, or a marble — reduces to the same three-step recipe.

Step 1: List or count all possible equally likely outcomes. This becomes the denominator. The word "equally likely" matters: it is only valid to count outcomes this way when each one is exactly as probable as every other, which is why a fair coin, a fair die, a well-shuffled deck, and a bag where any ball is picked "at random" are the standard settings.

Step 2: List or count the outcomes that satisfy the given condition. This becomes the numerator — only the outcomes that match exactly what the question describes.

Step 3: Apply the classical formula.

where: $P(E)$ is the probability of event $E$, the numerator counts outcomes matching $E$, and the denominator counts every equally likely outcome of the experiment. Almost every mistake in this topic happens in Step 1 or Step 2 — a wrong sample space, not a wrong formula. Four sample spaces are worth memorising cold:

• One coin: 2 outcomes — H, T.
• Two coins tossed together: 4 outcomes — HH, HT, TH, TT (note HT and TH are different outcomes, since it matters which coin shows what).
• One die: 6 outcomes — 1 through 6.
• Two dice thrown together: 36 outcomes — every ordered pair $(x,y)$ with $x, y \in \{1,\dots,6\}$, since the two dice act independently and $(2,5)$ is a different outcome from $(5,2)$.

A standard deck has 52 cards in 4 suits (spades, hearts, diamonds, clubs) of 13 cards each — Ace, 2 through 10, Jack, Queen, King. There are 26 red cards, 26 black cards, and 12 face cards total (Jack, Queen, King of each suit; the Ace is not a face card). For a bag of balls or marbles, the total outcome count is simply the total number of objects in the bag, since drawing "at random" guarantees each object is equally likely to be picked. For two-dice problems that ask about a sum, difference, or product, it is worth actually sketching a $6 \times 6$ grid of pairs on scratch paper — this single habit prevents the two most common errors in the whole chapter: missing a pair, or counting one twice.

The classical formula is not just a rule to memorise — it is a direct consequence of "equally likely." If an experiment has $n$ total equally likely outcomes, then by symmetry each single outcome carries probability $\tfrac{1}{n}$. An event $E$ made up of $k$ of those outcomes then has probability $k \times \tfrac{1}{n} = \tfrac{k}{n}$, which is exactly favourable outcomes over total outcomes. This is why the entire chapter is really about counting: once $n$ and $k$ are correctly identified, the "probability" step is a single division, not new mathematics. It also explains why $0 \le P(E) \le 1$ always: $k$ can be at most $n$ (probability 1, a certain event) and at least $0$ (probability 0, an impossible event), and why the complement rule $P(\bar E) = 1 - P(E)$ works — the favourable outcomes for $E$ and for "not $E$" together make up every outcome exactly once, so their counts, and hence their probabilities, must add to the whole, which is 1.

Problem: Two dice are thrown together. Find the probability that (a) the sum of the two numbers is 7, (b) the sum is more than 10, (c) a card drawn at random from a standard deck of 52 is a king, and (d) the card drawn is not an ace.

Step 1. Total outcomes for two dice: $6 \times 6 = 36$, since each die's six faces combine independently with the other die's six faces.

Step 2. List the pairs $(x,y)$ with $x + y = 7$: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ — 6 favourable outcomes. So $P(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6}$.

Step 3. For sum more than 10, the only possibilities are sum = 11 or sum = 12. Sum = 11 gives $(5,6), (6,5)$ — 2 outcomes; sum = 12 gives $(6,6)$ — 1 outcome. Total favourable $= 2 + 1 = 3$, so $P(\text{sum} > 10) = \frac{3}{36} = \frac{1}{12}$.

Step 4. For the card questions, total outcomes $= 52$. There are 4 kings in the deck (one per suit), so $P(\text{king}) = \frac{4}{52} = \frac{1}{13}$.

Step 5. There are 4 aces, so $P(\text{ace}) = \frac{4}{52} = \frac{1}{13}$. By the complement rule, $P(\text{not an ace}) = 1 - \frac{1}{13} = \frac{12}{13}$.

Final answers: $P(\text{sum}=7) = \frac{1}{6}$, $P(\text{sum}>10) = \frac{1}{12}$, $P(\text{king}) = \frac{1}{13}$, $P(\text{not an ace}) = \frac{12}{13}$.

Competitive stretch

Problem: A card is drawn at random from a standard deck. Find the probability it is a red face card.

Step 1. Face cards (Jack, Queen, King) number 3 per suit, 12 in total across all 4 suits. Red suits are hearts and diamonds, so red face cards number $3 \times 2 = 6$.

Step 2. $P(\text{red face card}) = \frac{6}{52} = \frac{3}{26}$. Answer: $\frac{3}{26}$.

• Adding instead of multiplying for two dice. The sample space for two dice has $6 \times 6 = 36$ outcomes, not $6 + 6 = 12$ — each die's reading combines with every reading of the other.
• Missing the mirror pair when listing sums. For "sum = 7," it is easy to list $(1,6)$ but forget $(6,1)$ — always scan systematically from $x=1$ to $x=6$, not just by feel.
• Confusing "face card" with "picture card" loosely, or forgetting the Ace is not a face card — the standard convention counts only Jack, Queen, King as face cards.
• Assuming equal likelihood without checking "at random." Outcomes are only equally likely because the problem says a coin/die/card/ball is fair or drawn at random — this phrase is what licenses the classical formula in the first place.

Ask the tutor: Stuck telling apart "sum equals a value" from "product equals a value" for two dice, or mixing up card categories — ask the AI tutor to build the full outcome table with you for your exact question.