PROBABILITY

Probability

The math is mostly arithmetic. The skill is reading the problem and knowing which of five rules it wants. Find the trigger word first — everything after that is calculation.

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The whole game

Find the Trigger → Grab the Rule

Before you touch any numbers: read the question, spot the trigger word, write down the formula. That pause is the skill.

See "or" / "either…or"

Addition Rule: P(A) + P(B) − P(A and B)

Don't forget to subtract the overlap — it's part of the formula, not optional.

See "neither…nor"

Complement of OR: 1 − P(A or B)

Find the OR answer first, then flip it. These two problems always come in pairs.

See "at least one"

Complement of None: 1 − P(none)

"None" is much easier to count than "at least one" — always use the flip.

See "and" / two independent events

Multiplication Rule: P(A) × P(B)

Independent = each event doesn't affect the other. Multiply the fractions.

See "given that" / "without replacement" / "if the first one was…"

Conditional: shrink the denominator (and maybe the numerator)

Total always drops by 1. Numerator drops by 1 only if the first draw matches the color/type you're asking about.

Worked Example 1 — Addition Rule (OR)

What is the probability that a card drawn from a standard deck is either a heart or a card with a number lower than 9?

1 · Trigger: "either…or" → Addition Rule. Write the formula before computing: P(A) + P(B) − P(A and B)

2 · P(heart) = 13/52. P(less than 9): cards 2–8 across all four suits = 7 values × 4 suits = 28 → P = 28/52 (Ace isn't a number card, so it's not "less than 9")

3 · Overlap — P(heart AND less than 9): hearts numbered 2–8 = 7 cards → 7/52

4 · 13/52 + 28/52 − 7/52 = 34/52

Don't skip the subtraction. Writing 13/52 + 28/52 = 41/52 double-counts the 7 hearts that are also less than 9. The "minus the overlap" is baked into the addition rule — not optional cleanup.

Worked Example 2 — Complement of OR (neither / nor)

What is the probability that the card is neither a heart nor a card with a number lower than 9?

1 · Trigger: "neither…nor" → Complement of OR → write: 1 − P(A or B)

2 · From WE1: P(heart or less than 9) = 34/52. Already done — no new setup needed.

3 · 1 − 34/52 = 18/52

Don't write 1 − P(heart) − P(less than 9). That's a different calculation and it's wrong — you'd double-subtract the overlap. The complement flips the whole OR answer, not each piece separately.

Worked Example 3 — Complement of None ("at least one")

If you roll six six-sided dice, what is the probability that at least one die shows a 4?

1 · Trigger: "at least one" → 1 − P(none). Counting all the ways to get one 4, two 4s, three 4s… is painful. Counting zero 4s is easy.

2 · P(no 4 on one die) = 5/6. Five of the six faces are not a 4.

3 · Six independent dice → multiply: P(no 4 on any die) = (5/6)⁶ = 15625/46656

4 · P(at least one 4) = 1 − 15625/46656 = 31031/46656

Two traps. (1) P(no 4) on one die is 5/6, not 1/6 — you want the faces that are NOT a 4. (2) Don't try P(one 4) = 1/6 multiplied by 6 to get 6/6 = 1. That overcounts and gives a number ≥ 1. The complement is the only clean path for "at least one."

Now you

Faded — Fill the Blanks

A jar holds 6 yellow, 9 purple, and 10 orange candies (25 total). You draw two candies without replacing the first one.

Sub-question A — fully worked (study this, don't solve it)

Find P(2nd candy is yellow | 1st candy was yellow)

1 · Trigger: "without replacing" → shrink the denominator. A yellow candy left the jar.

2 · Yellow count: 6 − 1 = 5 (first draw was yellow → yellow count shrinks)

3 · Total count: 25 − 1 = 24 (total always drops by 1, no exceptions)

4 · P(2nd yellow | 1st yellow) = 5/24

Sub-question B — fill in the blanks

Find P(2nd candy is orange | 1st candy was yellow)

1 · Trigger: "without replacing" → shrink the denominator. A yellow candy left the jar.

2 · Orange count: ______ (did removing a yellow candy change the orange count?)

3 · Total count: 25 − ______ = ______

4 · P(2nd orange | 1st yellow) = ______ / ______

Sub-question C — mostly blank, you drive

Find P(2nd candy is purple | 1st candy was purple)

1 · What left the jar on the first draw? ______

2 · Purple count: 9 − ______ = ______ (does the first draw match the color you're finding?)

3 · Total count: ______

4 · P(2nd purple | 1st purple) = ______ / ______

Check all three ›

A (worked for you): P = 5/24. Yellow shrinks (6 → 5), total shrinks (25 → 24).

B: Orange stays at 10 — you removed a yellow, not an orange. Total: 25 − 1 = 24. P(2nd orange | 1st yellow) = 10/24.

C: A purple candy left the jar. Purple: 9 − 1 = 8 (yes, it matches — so the numerator drops). Total: 24. P(2nd purple | 1st purple) = 8/24.

The rule in plain English: The denominator always drops by 1 — one candy left the jar. The numerator drops by 1 only if the first draw matches the color you're asking about.

Practice Set (6) — Name the Trigger First

For each problem, write the trigger word and the rule name before you compute. Work it on paper, then open the answer to check yourself.

1 · Single draw — King, diamond, diamond < 6 ›

Problem. A standard deck of 52 cards is shuffled. You draw one card at random. (a) What is the probability the card is a King? (b) What is the probability the card is a diamond? (c) What is the probability the card is a diamond with a number lower than 6? (Ace doesn't count as a number card.)

No special trigger — basic counting.

(a) 4 Kings in 52 cards. P(King) = 4/52.

(b) 13 diamonds. P(diamond) = 13/52.

2 · King or diamond ›

Problem. You draw one card from a standard deck. What is the probability it is either a King or a diamond?

Trigger: "either…or" → Addition Rule.

P(King) = 4/52 · P(diamond) = 13/52 · P(King AND diamond) = 1/52 (only the King of Diamonds is both).

P(King or diamond) = 4/52 + 13/52 − 1/52 = 16/52.

3 · 3 on first die and even on second ›

Problem. You roll two six-sided dice. What is the probability you roll a 3 on the first die and an even number on the second die?

Trigger: "and" + two dice (independent) → Multiplication Rule.

P(3 on first die) = 1/6. P(even on second die: 2, 4, 6) = 3/6.

P = 1/6 × 3/6 = 3/36 = 1/12.

4 · Neither King nor diamond ›

Problem. Same deck as #2. What is the probability the card is neither a King nor a diamond?

Trigger: "neither…nor" → Complement of OR.

From #2: P(King or diamond) = 16/52.

P(neither) = 1 − 16/52 = 36/52.

5 · At least one 6 in four dice ›

Problem. You roll four six-sided dice. What is the probability that at least one die shows a 6?

Trigger: "at least one" → 1 − P(none).

P(no 6 on one die) = 5/6. P(no 6 on any of 4 dice) = (5/6)⁴ = 625/1296.

P(at least one 6) = 1 − 625/1296 = 671/1296 ≈ 0.518.

6 · Marble bag — conditional & both red ›

Problem. A bag contains 4 red, 10 green, and 6 blue marbles (20 total). You draw two marbles without replacing the first. (a) What is P(2nd marble is red | 1st marble was red)? (b) What is P(2nd marble is blue | 1st marble was green)? (c) What is the probability both marbles are red?

Trigger: "without replacing" / "given" → shrink the denominator. (4 red, 10 green, 6 blue = 20 total)

(a) First draw removes a red: 3 red remain, 19 total. P(2nd red | 1st red) = 3/19.

(b) First draw removes a green: blue unchanged at 6, 19 total. P(2nd blue | 1st green) = 6/19.

Part (c) combines the multiplication rule with conditional — you need both triggers: "and" plus "without replacing."

Before You Leave — Add This to Your Sheet

Write the five triggers on your cheat sheet with the rule next to each — or → add and subtract overlap · neither/nor → flip the OR · at least one → flip the none · and → multiply · given / without replacement → shrink the denominator. Those five lines are the whole topic.

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