Applications of Probability
Unit 6 of Geometry. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Sample spaces, independence, conditional probability, two-way tables, the addition rule, using probability to make decisions.
How this unit is tested
Start by identifying what kind of probability question you're facing: is it about counting outcomes (sample space), reading a table (two-way table), combining two events with 'or' (addition rule), combining two events with 'and' given information (conditional probability or multiplication rule), or comparing expected outcomes to make a choice (decision making)? The wording is your clue — 'or' signals the addition rule, 'given that' or a restricted group signals conditional probability, and 'should you...' or 'is it worth it' signals expected value.
For two-way tables, always find the correct denominator first. A marginal probability uses the grand total; a conditional probability restricts you to a single row or column total, not the grand total. Write out P(A), P(B), and P(A and B) explicitly before combining them — most errors come from skipping this and guessing which numbers to divide.
To test independence, never assume it. Compute P(A|B) and compare it to P(A); if they match (or equivalently if P(A and B) = P(A)×P(B)), the events are independent. If they don't match, the events are dependent, and that dependence usually tells you something meaningful about the situation being modeled.
For decision-making problems, compute the expected value of each option (multiply each outcome by its probability and add) and compare. Expected value tells you what happens 'on average over many repetitions,' not what happens in any one trial — a game or policy can have a bad expected value even if it feels appealing for a single try, and that distinction is often the point of the question.
What you have to know
Sample Space and the Counting Principle
The sample space is the set of all possible outcomes of an experiment. If one stage of an experiment has m outcomes and a second independent stage has n outcomes, the combined sample space has m×n outcomes.
General Addition Rule
P(A or B) = P(A) + P(B) − P(A and B). If A and B are mutually exclusive (cannot both occur), P(A and B) = 0, so P(A or B) = P(A) + P(B).
Conditional Probability
P(A|B) = P(A and B) / P(B), read as 'the probability of A given that B has occurred.' It restricts the sample space to only the outcomes where B is true.
Independence
Events A and B are independent if P(A and B) = P(A)×P(B), equivalently if P(A|B) = P(A). Independence means knowing B occurred does not change the probability of A.
Expected Value for Decisions
For a random outcome X with possible values x₁, x₂, ... and probabilities P(x₁), P(x₂), ..., the expected value is E(X) = Σ xᵢ·P(xᵢ). Comparing expected values of two choices is a standard way to decide which is better 'in the long run.'
14 practice questions
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A spinner has 4 equally likely sections (red, blue, green, yellow) and a fair coin is flipped once. How many outcomes are in the combined sample space?
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Answer. 8 outcomes
By the counting principle, multiply the number of outcomes at each stage: 4 spinner outcomes × 2 coin outcomes = 8 total outcomes, such as (red, heads), (red, tails), and so on. -
Two standard six-sided dice are rolled. How many outcomes in the sample space give a sum of 8?
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Answer. 5 outcomes
List the ordered pairs summing to 8: (2,6), (3,5), (4,4), (5,3), (6,2). That is 5 outcomes out of the 36 total in the sample space. -
A single card is drawn from a standard 52-card deck. What is P(drawing a king or a queen)?
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Answer. 2/13 (or 8/52)
A card cannot be both a king and a queen, so these events are mutually exclusive: P(K or Q) = P(K) + P(Q) = 4/52 + 4/52 = 8/52 = 2/13. -
A single card is drawn from a standard deck. What is P(drawing a king or a heart)?
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Answer. 4/13 (or 16/52)
These events overlap (the king of hearts), so use the general addition rule: P(K or H) = P(K) + P(H) − P(K and H) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13. -
Which of the following pairs of events, when rolling one six-sided die, is best described as mutually exclusive?
- Rolling an even number and rolling a number greater than 4
- Rolling an even number and rolling an odd number
- Drawing a heart and drawing a face card from a deck
- Flipping heads on a coin and rolling a 3 on a die
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Answer. Rolling an even number and rolling an odd number
A single roll cannot be both even and odd, so P(even and odd) = 0, making them mutually exclusive. The other pairs share outcomes (for example, 6 is both even and greater than 4), so they are not mutually exclusive. -
Of 300 surveyed students, 150 play a sport and 150 do not; among sport players 60 have a job and 90 do not; among non-players 40 have a job and 110 do not. What is the probability a randomly chosen student does NOT play a sport?
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Answer. 0.5 (150/300)
This is a marginal probability, so use the grand total as the denominator: 150 non-players out of 300 total students gives 150/300 = 0.5. -
Using the same survey (150 sport / 150 no sport; sport-job 60, sport-no job 90; no sport-job 40, no sport-no job 110; total 300), what is the probability a student plays a sport AND has a job?
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Answer. 0.2 (60/300)
This is a joint probability, read directly from the interior cell where 'sport' and 'job' overlap: 60 students out of 300 total, giving 60/300 = 1/5 = 0.2. -
Using the same survey data, find P(has a job | plays a sport).
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Answer. 0.4 (60/150)
Conditioning on 'plays a sport' restricts the denominator to the sport row total (150), not the grand total. Of those 150 sport players, 60 have a job, so P(job|sport) = 60/150 = 0.4. -
Using the same survey data, find P(plays a sport OR has a job).
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Answer. ≈0.633 (190/300)
The events overlap, so use the addition rule: P(Sport)+P(Job)−P(Sport and Job) = 0.5 + 1/3 − 0.2 ≈ 0.633. This matches (300−110)/300, since 110 is the only group excluded (no sport, no job). -
Using the same survey data, are 'plays a sport' and 'has a job' independent events? Justify your answer with a calculation.
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Answer. No, they are not independent
Compare P(Sport|Job) to P(Sport): P(Sport|Job) = 60/100 = 0.6, while P(Sport) = 0.5. Since 0.6 ≠ 0.5, having a job changes the probability of playing a sport, so the events are dependent. -
Which of the following equations correctly defines two independent events A and B?
- P(A and B) = P(A) + P(B)
- P(A and B) = P(A) × P(B)
- P(A|B) = P(B)
- P(A and B) = 0
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Answer. P(A and B) = P(A) × P(B)
Independence means the occurrence of one event does not change the probability of the other, which is captured algebraically by P(A and B) = P(A)×P(B); equivalently P(A|B) = P(A). -
In a city, 30% of days are rainy. On rainy days, the probability of heavy traffic is 0.8. What is the probability that a randomly chosen day is both rainy and has heavy traffic?
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Answer. 0.24
Use the multiplication form of the conditional probability rule: P(A and B) = P(A) × P(B|A) = 0.3 × 0.8 = 0.24, where A = rainy day and B = heavy traffic. -
A carnival game costs $3 to play. You draw one marble from a bag of 4 red and 16 other marbles (20 total); a red marble wins $10, otherwise you win nothing. Based on expected value, should you expect to profit from playing many times?
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Answer. No — the expected value is a \$1 loss per play
P(red) = 4/20 = 0.2, so expected winnings = 0.2 × \$10 = \$2. Subtracting the \$3 cost gives a net expected value of \$2 − \$3 = −\$1 per play, meaning the game is unfavorable in the long run even though a single play could still win \$10. -
An extended warranty costs $50. There is a 10% chance the covered item fails and needs a $300 repair, and a 90% chance nothing goes wrong. Based purely on expected value, is buying the warranty or skipping it the lower-cost choice on average?
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Answer. Skipping the warranty has the lower expected cost ($30 vs. $50)
Without the warranty, expected cost = P(fail)×$300 = 0.10×$300 = $30. With the warranty, the cost is a guaranteed $50. Since $30 < $50, skipping the warranty minimizes expected cost, though a customer might still buy it to avoid the risk of a large unexpected repair bill.
What people get wrong
- Treating independent and mutually exclusive as the same idea. Mutually exclusive events (P(A and B)=0) are actually maximally dependent — knowing one occurred tells you the other definitely did not. Always check the correct definition for the situation before applying a rule.
- Using the grand total as the denominator when the question asks for a conditional probability. If the question says 'given that' or 'among students who...', the denominator must be the row or column total for that condition, not the full sample size.
- Forgetting to subtract the overlap in the addition rule when the events are not mutually exclusive, which double-counts outcomes in both A and B. Before adding P(A) + P(B), always ask whether an outcome could satisfy both events at once.
- Reversing the order in conditional probability, computing P(B|A) when the question asks for P(A|B). Write P(A|B) = P(A and B)/P(B) explicitly and plug in the matching quantities rather than dividing whichever two numbers seem convenient.
- Confusing 'expected value favors playing' with 'you will win.' Expected value describes the long-run average outcome; a negative expected value means the choice is unfavorable on average even though a single trial could still win.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of Geometry. Free, and no account needed to start.