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Probability of Simple and Compound Events

Find the chance of one event, of two events together (and), of either event (or), with and without replacement.

Practice this skill 25 min Probability of simple and compound events Leer en español
Key idea

What you'll learn

In this lesson you will learn how to find the probability — the chance — that something happens. You will start with a single event (one spin, one draw) and build up to compound events: two or more things happening together ("and"), either of two things happening ("or"), and draws with or without replacement.

Why it matters on the GED: probability questions show up as spinners, bags of marbles, dice, survey tables and real situations like quality control or weather forecasts. You may be asked for an answer as a fraction, a decimal or a percent, so you need to move between all three.

Real life: a 30% chance of rain, the odds of a defective part in a shipment, the chance of winning a raffle, or the chance that two random customers both pay with cash — these are all probability.

Words to know

  • Experiment — an action with uncertain results. Example: rolling a die.
  • Outcome — one possible result. Example: rolling a 4.
  • Sample space — the list of all possible outcomes. Example: {1, 2, 3, 4, 5, 6} for one die.
  • Event — the outcome(s) you are interested in. Example: "roll an even number" = {2, 4, 6}.
  • Favorable outcomes — the outcomes that make the event happen. Example: 3 favorable outcomes for "even".
  • Probability P(A) — a number from 0 (impossible) to 1 (certain). Example: P(\text{even}) = \tfrac{3}{6} = \tfrac{1}{2}.
  • Complement ("not A") — everything that is not A. Example: P(\text{not even}) = 1 - \tfrac{1}{2} = \tfrac{1}{2}.
  • Compound event — two or more events combined with "and" or "or".
  • Independent events — one doesn't change the chance of the other. Example: flipping a coin and rolling a die.
  • Dependent events — the first result changes the chance of the second. Example: drawing two cards without putting the first back.
  • With replacement — the item is put back before the next draw (the total stays the same).
  • Without replacement — the item is kept out (the total goes down by 1).
  • Mutually exclusive — the events can't both happen at once. Example: rolling a 2 and rolling a 5 on one roll.

All Math words in the glossary

1. Simple probability

When all outcomes are equally likely:

P(\text{event}) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}

Example: A bag has 5 red, 3 blue and 2 green marbles (10 in all). P(\text{blue}) = \dfrac{3}{10} = 0.3 = 30\%.

A bag holding 10 marbles: 5 red (R), 3 blue (B) and 2 green (G). The probability of drawing blue is 3 out of 10 = 0.3 = 30%.Bag of 10 marblesRRRRRBBBGGR = red, B = blue, G = greenRed: 5Blue: 3Green: 2Total: 10P(blue) = 3/10= 0.3 = 30%

Example: The spinner below has 8 equal sections. P(\text{a number greater than 5}) = \dfrac{3}{8} because 6, 7 and 8 are favorable.

A spinner divided into 8 equal sections numbered 1, 2, 3, 4, 5, 6, 7, 8, with an arrow at the center.12345678

Writing the answer. Any of these forms may be asked for: \tfrac{3}{8} = 0.375 = 37.5\%. To convert a fraction to a percent, divide and multiply by 100.

The complement rule. The chance something does not happen is 1 minus the chance it does:

P(\text{not } A) = 1 - P(A)

Example: If there is a 30% chance of rain, there is a 100\% - 30\% = 70\% chance of no rain.

Probability vs. odds. Probability compares favorable to total. "Odds" compare favorable to unfavorable (3 blue to 7 not-blue). The GED asks for probability, so always divide by the total.

From data. In a table or survey, the "total" is the number of people or items in the group you are choosing from.

Example: 40 employees: 12 drive, 18 take the bus, 10 walk. P(\text{walk}) = \tfrac{10}{40} = \tfrac{1}{4}.

2. "And": independent and dependent events

For "A and B" (both happen), multiply:

P(A \text{ and } B) = P(A) \times P(B \text{, after } A)

Independent events — the second probability doesn't change.

Example: Roll a die and flip a coin. P(4 \text{ and heads}) = \dfrac{1}{6} \times \dfrac{1}{2} = \dfrac{1}{12}.

Sample-space grid for rolling a die (1 to 6, columns) and flipping a coin (H or T, rows): 12 equally likely outcomes. The single cell "4, H" is highlighted, so P(4 and heads) = 1/12, which equals 1/6 times 1/2.DieCoin123456H1, H2, H3, H4, H5, H6, HT1, T2, T3, T4, T5, T6, T6 × 2 = 12 equally likely outcomes (H = heads, T = tails)P(4 and heads) = 1/12 = 1/6 × 1/2

Example (with replacement): A bag has 4 red and 6 blue marbles. Draw one, put it back, draw again. P(\text{red, red}) = \dfrac{4}{10} \times \dfrac{4}{10} = \dfrac{16}{100} = 0.16.

Dependent events — the first draw changes what is left. Update both the favorable count and the total.

Example (without replacement): Same bag, but keep the first marble. After a red is drawn, 3 red remain out of 9 marbles. P(\text{red, red}) = \dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15} \approx 0.133.

Tree diagram for drawing two marbles without replacement from 4 red and 6 blue. First draw: red 4/10, blue 6/10. After a red: red 3/9, blue 6/9. After a blue: red 4/9, blue 5/9. Outcomes: red-red 12/90, red-blue 24/90, blue-red 24/90, blue-blue 30/90. The red-red path is highlighted: 4/10 times 3/9 = 12/90 = 2/15, about 0.133.1st draw2nd drawOutcomeStart4/10Red3/9Redred, red12/906/9Bluered, blue24/906/10Blue4/9Redblue, red24/905/9Blueblue, blue30/90P(red, red) = 4/10 × 3/9 = 12/90 = 2/15 ≈ 0.133
1st draw2nd drawP(\text{red, red})
With replacement\tfrac{4}{10}\tfrac{4}{10}0.16
Without replacement\tfrac{4}{10}\tfrac{3}{9}\approx 0.133

Notice that "and" always makes the probability smaller (or equal) — it is harder for two things to happen than one.

"At least one" — use the complement. P(\text{at least one}) = 1 - P(\text{none}).

Example: Each part has a 10% chance of being defective. For 3 parts, P(\text{none defective}) = 0.9^3 = 0.729, so P(\text{at least one defective}) = 1 - 0.729 = 0.271.

3. "Or": adding probabilities

For "A or B" (at least one of them happens on a single trial), add — but don't count anything twice:

P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)

Mutually exclusive events (they can't both happen): the overlap is 0, so just add.

Example: 40 employees: 12 drive, 18 bus, 10 walk (each person uses one way). P(\text{drive or walk}) = \dfrac{12}{40} + \dfrac{10}{40} = \dfrac{22}{40} = 0.55.

Overlapping events subtract the overlap once.

Example: In a class of 30, 18 play a sport, 12 are in the band, and 5 do both. Adding 18 + 12 = 30 counts those 5 people twice.

P(\text{sport or band}) = \frac{18}{30} + \frac{12}{30} - \frac{5}{30} = \frac{25}{30} = \frac{5}{6}
Venn diagram for a class of 30. Sport circle (18) and Band circle (12) overlap in 5 students who do both. Sport only 13, band only 7, neither 5. Sport or band = 13 + 5 + 7 = 25, which equals 18 + 12 - 5, so P = 25/30 = 5/6.Class of 30Sport (18)Band (12)135both7neither: 513 + 5 + 7 = 25 = 18 + 12 − 5 (the 5 counted once)P(sport or band) = 25/30 = 5/6

Two-way tables. Read the row or column that matches the group you are choosing from.

CoffeeTeaTotal
Under 30384280
30 and over9228120
Total13070200
  • Pick anyone: P(\text{tea}) = \tfrac{70}{200} = 0.35.
  • Pick anyone: P(\text{under 30 and tea}) = \tfrac{42}{200} = 0.21.
  • Pick from the tea drinkers only: P(\text{under 30}) = \tfrac{42}{70} = 0.6. The total is now 70, not 200.

Decision guide: "and" / "both" / "in a row" → multiply. "or" / "either" → add (minus the overlap). "not" / "at least one" → use 1 - P.

Formula

Formulas (memorize these)

P(A) = \frac{\text{favorable outcomes}}{\text{total outcomes}} \qquad P(\text{not } A) = 1 - P(A) P(A \text{ and } B) = P(A) \times P(B \text{ after } A) P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)
  • P(A) is always between 0 and 1 (0% to 100%).
  • For independent events, P(B \text{ after } A) is just P(B).
  • For mutually exclusive events, P(A \text{ and } B) = 0.

These probability rules are not on the GED formula sheet — memorize them. The good news: they all come from one idea, favorable ÷ total.

Worked example

Worked example 1 (easy): a single draw

A jar holds 6 cherry, 4 lemon and 5 grape candies. Mia takes one without looking. What is the probability it is lemon?

  1. Total candies: 6 + 4 + 5 = 15.
  2. Favorable (lemon): 4.
  3. P(\text{lemon}) = \dfrac{4}{15} \approx 0.267.
A jar of 15 candies: 6 cherry (C), 4 lemon (L) and 5 grape (G). Lemon is the favorable outcome: P(lemon) = 4/15, about 0.267 or 26.7%.Jar of 15 candiesCCCCCCLLLLGGGGGC = cherry, L = lemon, G = grapeCherry: 6Lemon: 4 ← favorableGrape: 5Total: 15P(lemon) = 4/15≈ 0.267 ≈ 26.7%

Answer: \tfrac{4}{15} (about 26.7%).

Check: P(\text{not lemon}) = \tfrac{11}{15}, and \tfrac{4}{15} + \tfrac{11}{15} = 1 ✓

Worked example

Worked example 2: independent "and"

A basketball player makes 70% of her free throws. Each shot is independent. What is the probability she makes both of her next two free throws?

  1. "Both" means "and" → multiply.
  2. P(\text{make, make}) = 0.7 \times 0.7 = 0.49.
Tree diagram for two independent free throws, each made with probability 0.7 and missed with 0.3. Outcomes: make-make 0.49, make-miss 0.21, miss-make 0.21, miss-miss 0.09 (total 1). The make-make path is highlighted: 0.7 times 0.7 = 0.49.1st shot2nd shotOutcomeStart0.7Make0.7Makemake, make0.490.3Missmake, miss0.210.3Miss0.7Makemiss, make0.210.3Missmiss, miss0.09P(make both) = 0.7 × 0.7 = 0.49 = 49%

Answer: 0.49, or 49%.

Check: Both made (0.49) is less than one made (0.7), which makes sense — "and" is harder ✓

Worked example

Worked example 3 (real world): without replacement

A box of 12 light bulbs contains 3 defective bulbs. An inspector picks 2 bulbs at random without replacement. What is the probability both are defective?

  1. First bulb: P(\text{defective}) = \dfrac{3}{12}.
  2. One defective bulb is gone: 2 defective out of 11 bulbs left. P(\text{2nd defective}) = \dfrac{2}{11}.
  3. Multiply: \dfrac{3}{12} \times \dfrac{2}{11} = \dfrac{6}{132} = \dfrac{1}{22}.
Tree diagram for picking 2 of 12 bulbs (3 defective, 9 good) without replacement. First bulb: defective 3/12, good 9/12. After a defective: defective 2/11, good 9/11. After a good: defective 3/11, good 8/11. Outcomes: both defective 6/132, defective then good 27/132, good then defective 27/132, both good 72/132. The both-defective path is highlighted: 3/12 times 2/11 = 6/132 = 1/22.1st bulb2nd bulbOutcomeStart3/12Defect2/11Defectdefective ×26/1329/11Gooddef., good27/1329/12Good3/11Defectgood, def.27/1328/11Goodgood ×272/132P(both defective) = 3/12 × 2/11 = 6/132 = 1/22

Answer: \tfrac{1}{22} (about 4.5%).

Check: Using combinations: \dfrac{{}_3C_2}{{}_{12}C_2} = \dfrac{3}{66} = \dfrac{1}{22} ✓

Worked example

Worked example 4: "or" with overlap

A survey of 50 shoppers found that 30 use coupons, 20 shop online, and 8 do both. A shopper is chosen at random. What is the probability the shopper uses coupons or shops online?

  1. "Or" → add, then subtract the overlap so the 8 aren't counted twice.
  2. P = \dfrac{30}{50} + \dfrac{20}{50} - \dfrac{8}{50} = \dfrac{42}{50}.
  3. \dfrac{42}{50} = 0.84.
Venn diagram for 50 shoppers. Coupons circle (30) and Online circle (20) overlap in 8 who do both. Coupons only 22, online only 12, neither 8. Coupons or online = 22 + 8 + 12 = 42, which equals 30 + 20 - 8, so P = 42/50 = 0.84.50 shoppersCoupons (30)Online (20)228both12neither: 822 + 8 + 12 = 42 = 30 + 20 − 8P(coupons or online) = 42/50 = 0.84

Answer: 0.84, or 84%.

Check: Coupons only 30 - 8 = 22, online only 20 - 8 = 12, both 8: 22 + 12 + 8 = 42 ✓

Worked example

Worked example 5 (GED-hard): work backwards

A bag has 3 green marbles and some white marbles. Two marbles are drawn with replacement. The probability that both are green is \tfrac{1}{16}. How many white marbles are in the bag?

  1. With replacement, P(\text{green, green}) = P(\text{green})^2 = \tfrac{1}{16}.
  2. So P(\text{green}) = \sqrt{\tfrac{1}{16}} = \tfrac{1}{4}.
  3. \tfrac{3}{\text{total}} = \tfrac{1}{4}, so the total is 3 \times 4 = 12 marbles.
  4. White marbles = 12 - 3 = 9.
A bag of 12 marbles: 3 green (G) and 9 white (W). Since P(green, green) with replacement is 1/16, P(green) = 1/4, so the total is 12 and there are 12 - 3 = 9 white marbles. Check: 3/12 times 3/12 = 9/144 = 1/16.Bag: 3 green + 9 whiteGGGWWWWWWWWWP(green, green) = 1/16P(green) = √(1/16) = 1/43 ÷ total = 1/4 → total = 12white = 12 − 3 = 9Check: 3/12 × 3/12 = 9/144= 1/16 ✓G = green, W = white

Answer: 9 white marbles.

Check: \tfrac{3}{12} \times \tfrac{3}{12} = \tfrac{9}{144} = \tfrac{1}{16} ✓

Calculator tip

Calculator tip: TI-30XS MultiView

  • Enter fractions with the n/d key: 3 n/d 12 ▶ × 2 n/d 11 enter gives \tfrac{1}{22} already simplified.
  • Switch forms with the toggle key: it flips the answer between fraction and decimal (\tfrac{1}{22} \leftrightarrow 0.0454…).
  • Powers for repeated "and": 0.9^3 → 0.9 ^ 3 enter = 0.729.
  • Percent: multiply the decimal by 100.

When NOT to use it: simple probabilities like \tfrac{3}{10} or \tfrac{1}{6} \times \tfrac{1}{2} are faster by hand. The calculator can't tell you whether to add or multiply, or whether the total should drop by 1 — set up the problem first.

Common mistake

Common mistakes

  • Giving odds instead of probability. 3 blue out of 10 marbles: odds are 3 : 7, but the probability is \tfrac{3}{10} (divide by the total), not \tfrac{3}{7}.
  • Adding for "and". P(4 \text{ and heads}) is \tfrac{1}{6} \times \tfrac{1}{2} = \tfrac{1}{12}, not \tfrac{1}{6} + \tfrac{1}{2} = \tfrac{2}{3}. Adding for "and" can even give answers bigger than 1!
  • Forgetting to update after "without replacement." After drawing a red, there are 1 fewer reds and 1 fewer marbles: \tfrac{4}{10} \times \tfrac{3}{9}, not \tfrac{4}{10} \times \tfrac{4}{10} (and not \tfrac{4}{10} \times \tfrac{3}{10}).
  • Double-counting the overlap in "or". 18 in sports + 12 in band is not 30 people if 5 do both. Subtract the overlap once: 25.
  • Confusing "at least one" with "all". P(\text{at least one defective}) = 1 - P(\text{none}), not 0.1 \times 0.1 \times 0.1.
  • Using the wrong total in a table. "Of the tea drinkers, what fraction…" means divide by the tea total (70), not the grand total (200).
Test strategy

GED test strategy

  • Sanity check: a probability must be between 0 and 1. Eliminate any answer greater than 1 or less than 0.
  • "And" can't be larger than either single probability; "or" can't be smaller than either one. Use this to eliminate choices quickly.
  • Read the question twice for the words replace / put back / without replacement, both, either, at least one, and of the ___ (which total to use).
  • Check the answer format: fraction, decimal or percent? Convert at the end, and round only at the end.
  • Simplify fractions — answer choices are usually in lowest terms. The n/d key does this for you.
  • Time: most probability questions take 1–2 minutes. If a problem looks long, write the two fractions and multiply.

Check yourself

Probability of simple and compound events

A bag contains 7 yellow, 4 green, and 9 purple tiles. One tile is drawn at random.

What is the probability that the tile is green?

Answer choices
Hint

Probability compares the number of green tiles to the number of tiles in the whole bag.

Check yourself

Probability of simple and compound eventsFill in

The spinner below has 8 equal sections.

A spinner divided into 8 equal sections numbered 1, 2, 3, 4, 5, 6, 7, 8, with an arrow at the center.12345678

The arrow is spun once. What is the probability that it lands on a number greater than 2? Write your answer as a decimal.

Type your answer in the box. Fractions like 3/4 are fine.

Hint

List the numbers on the spinner that are greater than 2 (be careful: is 2 itself included?).

Summary

  • P(A) = \dfrac{\text{favorable}}{\text{total}}, always from 0 to 1; P(\text{not } A) = 1 - P(A).
  • "And" → multiply. Independent / with replacement: the second fraction is unchanged. Dependent / without replacement: subtract 1 from the favorable count and the total.
  • "Or" → add, then subtract the overlap: P(A) + P(B) - P(A \text{ and } B).
  • "At least one" → 1 - P(\text{none}).
  • In tables, divide by the total of the group you are choosing from. None of these rules are on the formula sheet.
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