Biostatistics Quiz: Basic Probability Rules

20 questions · exam conditions

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Basic Probability RulesQuestion 1 of 20

Given $P(A)=0.5, P(B)=0.4, P(A\cap B)=0.2$ , find $P(A|B')$ .

0.30

0.50

0.60

0.75

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Biostatistics Quiz

Biostatistics Quiz: Basic Probability Rules

Practice Basic Probability Rules in Biostatistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Basic Probability Rules, giving you a quick way to practice the rules, question types, and explanations that matter most for Biostatistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Given $P(A)=0.5, P(B)=0.4, P(A\cap B)=0.2$ , find $P(A|B')$ .

0.30
0.50 (correct answer)
0.60
0.75

Explanation: Since B' has probability 0.6, and A∩B' has probability P(A) - P(A∩B) = 0.5 - 0.2 = 0.3, divide 0.3 by 0.6 to get 0.50. The tempting error is 0.75, which comes from using P(B) = 0.4 as the denominator instead of P(B').

Question 2

If $P(A\cup B)=0.6, P(B)=0.3, P(A|B)=0.2$ , find $P(A)$ .

0.06
0.24
0.30
0.36 (correct answer)

Explanation: First find the intersection: P(A|B) times P(B) = 0.2 × 0.3 = 0.06. Then use the union formula: P(A∪B) = P(A) + P(B) - P(A∩B), so 0.6 = P(A) + 0.3 - 0.06, giving P(A) = 0.36. The tempting mistake is treating 0.06 as the answer, but that is only the intersection, not P(A).

Question 3

Test +: 90% if diseased, 5% if healthy; prevalence 1%. Find $P(D|+)$ .

0.009
0.154 (correct answer)
0.167
0.900

Explanation: Out of 1000 people, 10 have disease and 990 are healthy. Of the 10 diseased, 90% test positive: 9. Of the 990 healthy, 5% test positive: 49.5. Total positive tests are 58.5, so the chance a positive test is truly diseased is 9 divided by 58.5, which equals 0.154. The tempting wrong answer 0.009 is the joint probability of being diseased and testing positive, not the conditional probability given a positive test.

Question 4

If A and B are independent, $P(A)=0.6, P(A\cup B)=0.8$ , find $P(B|A)$ .

0.200
0.300
0.500 (correct answer)
0.800

Explanation: For independent events, P(A and B) = P(A)P(B), so P(A or B) = P(A) + P(B) - P(A)P(B). Plug in 0.6 + P(B) - 0.6P(B) = 0.8, giving P(B) = 0.5. Independence means P(B|A) = P(B), so the answer is 0.500. The tempting wrong value 0.300 is P(A and B), not the conditional probability; remember to divide by P(A).

Question 5

Patients get Drug A or B: $P(A)=0.3, P(B)=0.7, P(R|A)=0.8, P(R|B)=0.6$ . Find $P(R)$ .

0.60
0.66 (correct answer)
0.70
0.74

Explanation: Weight each conditional probability by the probability of receiving that drug: 0.3 times 0.8 gives 0.24, and 0.7 times 0.6 gives 0.42. Adding these gives 0.66. A tempting mistake is to pick 0.60, but that ignores the 30% of patients treated with Drug A.

Question 6

In a study, 40% have A, 30% have B, and 50% of those with B have A. Find $P(A∪B)$ .

0.70
0.15
0.55 (correct answer)
0.50

Explanation: Since 50% of those with B have A, the overlap P(A and B) is 0.5 x 0.3 = 0.15. Add the individual probabilities and subtract the overlap: 0.40 + 0.30 - 0.15 = 0.55. The tempting 0.70 simply sums P(A) and P(B) and forgets that A and B overlap.

Question 7

If $P(A)=0.5$ and $P(B|A)=0.3$ , find the probability of A and not B.

0.35 (correct answer)
0.70
0.15
0.30

Explanation: Start with P(A and B) = P(B|A) * P(A) = 0.3 * 0.5 = 0.15. Since P(A) is split into P(A and B) plus P(A and not B), subtract: 0.5 - 0.15 = 0.35. The tempting 0.15 is the probability of A and B, not A and not B.

Question 8

If $P(A)=0.6$ , $P(B)=0.5$ , and $P(A∪B)=0.8$ , find $P(A|B)$ .

0.30
0.60 (correct answer)
0.80
0.50

Explanation: Use the addition rule to find the intersection: 0.8 = 0.6 + 0.5 minus P(A and B), so P(A and B) = 0.3. Then P(A|B) = P(A and B) divided by P(B) = 0.3 / 0.5 = 0.60. The tempting 0.30 is only the intersection, not the conditional probability.

Question 9

30% get drug, 70% placebo. Response is 90% with drug, 50% with placebo. Let D=drug, R=response. Find $P(D|R)$ .

0.900
0.270
0.620
0.435 (correct answer)

Explanation: Among 100 people, 30 get drug and 27 respond. 70 get placebo and 35 respond, so 62 respond total. The 62 responders include 27 who took the drug, so 27/62 = 0.435. The tempting wrong answer is 0.270, which is P(drug and response), not P(drug given response).

Question 10

$A$ and $B$ are independent, $P(A)=0.7$ and $P(B)=0.4$ . Find the probability that exactly one occurs.

0.82
0.28
0.18
0.54 (correct answer)

Explanation: Exactly one occurs as A without B or B without A. With independence, multiply: 0.7 * 0.6 = 0.42 and 0.3 * 0.4 = 0.12, then add to get 0.54. The tempting 0.82 is P(A or B), which also counts the chance both happen, so it is too large for exactly one.

Question 11

In a population study, the probability of having gene variant A is 0.25, gene variant B is 0.40, and having neither variant is 0.50. What is the probability of having both variants A and B?

0.10
0.15 (correct answer)
0.20
0.25
0.65

Explanation: When you encounter probability questions involving multiple conditions, think systematically about the relationships between events using the fundamental probability rules. Let's define our events: A = having gene variant A, B = having gene variant B. We know P(A) = 0.25, P(B) = 0.40, and P(neither A nor B) = 0.50. Since having "neither variant" means not having A AND not having B, we can write this as P(A' ∩ B') = 0.50. Using the complement rule, if the probability of having neither variant is 0.50, then the probability of having at least one variant is P(A ∪ B) = 1 - 0.50 = 0.50. Now apply the addition rule: P(A ∪ B) = P(A) + P(B) - P(A ∩ B). Substituting our known values: 0.50 = 0.25 + 0.40 - P(A ∩ B), which gives us P(A ∩ B) = 0.65 - 0.50 = 0.15. Choice A (0.10) likely comes from incorrectly multiplying P(A) × P(B) assuming independence, but these events aren't independent. Choice C (0.20) might result from miscalculating 1 - 0.50 - 0.25 - 0.40, incorrectly treating the probabilities as mutually exclusive. Choice D (0.25) could come from confusing P(A ∩ B) with P(A). Study tip: For overlapping probability problems, always check if the individual probabilities and intersections add up logically using the addition rule. Draw Venn diagrams when possible to visualize the relationships between events.

Question 12

In a genetics study, P(brown eyes) = 0.70, P(brown hair) = 0.60, and P(brown eyes AND brown hair) = 0.50. Given that a person has brown hair, what is the probability they do NOT have brown eyes?

0.167 (correct answer)
0.300
0.333
0.500
0.833

Explanation: When you encounter conditional probability questions in biostatistics, you're dealing with how one event affects the probability of another. This question asks for P(NOT brown eyes | brown hair), which requires careful application of conditional probability formulas. Start with the conditional probability formula: P(A|B) = P(A AND B) / P(B). Here, you want P(NOT brown eyes | brown hair). Since P(brown eyes | brown hair) + P(NOT brown eyes | brown hair) = 1, you can find P(brown eyes | brown hair) first, then subtract from 1. Using the given information: P(brown eyes | brown hair) = P(brown eyes AND brown hair) / P(brown hair) = 0.50 / 0.60 = 0.833. Therefore, P(NOT brown eyes | brown hair) = 1 - 0.833 = 0.167. Looking at the wrong answers: B) 0.300 represents P(NOT brown eyes) in the general population (1 - 0.70), ignoring the conditional aspect entirely. C) 0.333 might result from incorrectly calculating 0.20/0.60, where 0.20 comes from mistakenly using P(brown hair) - P(brown eyes AND brown hair). D) 0.500 simply uses the intersection probability without proper conditional calculation. The key trap here is confusing unconditional probability with conditional probability. Always remember that conditioning on additional information changes the sample space. When you see "given that" in probability questions, immediately think conditional probability and use P(A|B) = P(A AND B) / P(B). Practice identifying what becomes your new denominator when the condition is applied.

Question 13

A pharmaceutical company tests drug efficacy using two methods independently. Method 1 correctly identifies effective drugs 92% of the time. Method 2 correctly identifies effective drugs 88% of the time. What is the probability that both methods fail to identify an effective drug?

0.0096 (correct answer)
0.0400
0.0800
0.1600
0.2000

Explanation: When you encounter questions about multiple diagnostic tests or methods working independently, you're dealing with probability calculations involving independent events. The key insight is recognizing what "both methods fail" means and how to calculate it. To find the probability that both methods fail to identify an effective drug, you need the failure rate for each method. Method 1 succeeds 92% of the time, so it fails 8% of the time (1 - 0.92 = 0.08). Method 2 succeeds 88% of the time, so it fails 12% of the time (1 - 0.88 = 0.12). Since the methods work independently, you multiply their individual failure probabilities:

P(\text{both fail}) = 0.08 \times 0.12 = 0.0096

Looking at the wrong answers: Choice B (0.0400) represents a common error where students might incorrectly add the failure rates (0.08 + 0.12 = 0.20) then try some manipulation. Choice C (0.0800) is simply the failure rate of Method 1 alone, suggesting the student ignored Method 2 entirely. Choice D (0.1600) might result from incorrectly squaring one of the success rates or making an arithmetic error in the multiplication. The correct answer is A (0.0096). Remember this pattern: for independent events, multiply probabilities. When calculating failure rates, always subtract the success rate from 1 first. This type of question frequently appears in biostatistics because diagnostic testing often involves multiple independent methods, and understanding combined failure rates is crucial for clinical decision-making.

Question 14

Two treatments are tested independently on different patient groups. Treatment A succeeds 75% of the time, Treatment B succeeds 80% of the time. Assuming the treatments work independently when combined, what is the probability that at least one treatment succeeds when both are used?

0.550
0.600
0.775
0.950 (correct answer)
1.550

Explanation: When you encounter probability questions about "at least one" success occurring, the most efficient approach is to use the complement rule: calculate the probability that neither treatment succeeds, then subtract from 1. Let's work through this step-by-step. Treatment A succeeds 75% of the time, so it fails 25% of the time (0.25). Treatment B succeeds 80% of the time, so it fails 20% of the time (0.20). Since the treatments work independently, the probability that both fail is:

0.25 \times 0.20 = 0.05

Therefore, the probability that at least one treatment succeeds is:

1 - 0.05 = 0.95

Looking at the wrong answers: Choice A (0.550) likely comes from incorrectly multiplying the success rates:

0.75 \times 0.80 = 0.60

, then making an additional error. Choice B (0.600) results from multiplying the individual success probabilities, which gives you the probability that both treatments succeed, not at least one. Choice C (0.775) appears to come from simply averaging the two success rates:

(0.75 + 0.80)/2 = 0.775

, which has no basis in probability theory. The correct answer is D (0.950). Remember this key strategy: when you see "at least one" in probability questions, think complement rule first. It's almost always easier to calculate the probability of the opposite event (none succeed) and subtract from 1, rather than calculating all the ways at least one can succeed directly.

Question 15

Two independent genetic mutations occur with probabilities 0.12 and 0.08 respectively. Given that at least one mutation occurred, what is the probability that both mutations occurred?

0.0096
0.048
0.051 (correct answer)
0.200
0.400

Explanation: When you encounter a question asking for the probability of an event "given that" another event occurred, you're dealing with conditional probability. This is a fundamental concept in biostatistics, especially when analyzing disease occurrence or genetic mutations. This problem requires using the conditional probability formula:

P(A|B) = \frac{P(A \cap B)}{P(B)}

. Here, you need to find the probability that both mutations occurred given that at least one occurred. Let's define the events: Mutation 1 occurs with

P(M_1) = 0.12

and Mutation 2 occurs with

P(M_2) = 0.08

. Since the mutations are independent,

P(\text{both}) = 0.12 \times 0.08 = 0.0096

. The probability of at least one mutation is:

P(\text{at least one}) = 1 - P(\text{neither}) = 1 - (1-0.12)(1-0.08) = 1 - 0.88 \times 0.92 = 1 - 0.8096 = 0.1904

. Therefore:

P(\text{both}|\text{at least one}) = \frac{0.0096}{0.1904} = 0.0504 \approx 0.051

Answer A (0.0096) is the probability of both mutations occurring without the condition - this ignores the "given that" clause. Answer B (0.048) likely results from calculation errors in the denominator. Answer D (0.200) suggests incorrectly assuming equal likelihood of the four possible outcomes. Remember: conditional probability questions always require dividing the intersection probability by the conditioning event's probability. Never ignore the "given that" information - it fundamentally changes the sample space you're working with.

Question 16

A screening program tests for three diseases independently. Disease A is detected 90% of the time when present, Disease B is detected 85% of the time, and Disease C is detected 95% of the time. What is the probability that all three tests fail to detect their respective diseases when all diseases are present?

0.000075 (correct answer)
0.000625
0.007125
0.100000
0.270000

Explanation: When you encounter questions about independent events in biostatistics, remember that "independent" means the outcome of one test doesn't affect the others, so you multiply probabilities together. Here, you need the probability that all three tests fail when all diseases are present. First, convert the detection rates to failure rates:

Disease A: 90% detection means 10% failure rate (0.10)
Disease B: 85% detection means 15% failure rate (0.15)
Disease C: 95% detection means 5% failure rate (0.05)

Since the tests are independent, multiply these failure probabilities:

P(\text{all tests fail}) = 0.10 \times 0.15 \times 0.05 = 0.000075

Looking at the wrong answers: B (0.000625) likely comes from incorrectly using 0.05 × 0.05 × 0.25, perhaps misreading one of the detection rates. C (0.007125) might result from adding the failure rates instead of multiplying them, or making decimal errors in calculation. D (0.100000) represents just the failure rate of Disease A alone, ignoring the other two diseases entirely. The key insight is that when all diseases are present simultaneously, the chance of missing every single one becomes very small because you're multiplying small probabilities together. This is why comprehensive screening programs are effective—even if individual tests aren't perfect, the probability of missing everything drops dramatically. Study tip: For independent events, always multiply probabilities. Convert percentages to decimals carefully, and remember that "failure to detect" = 1 - detection rate.

Question 17

In a medical database, 35% of patients have condition X, 28% have condition Y, and 12% have both conditions. If a patient does not have condition X, what is the probability that they have condition Y?

0.123
0.160
0.246 (correct answer)
0.280
0.400

Explanation: When you encounter conditional probability questions in biostatistics, you're dealing with how one event's probability changes when you know another event has or hasn't occurred. The key formula is

P(A|B) = \frac{P(A \cap B)}{P(B)}

. Here, you need

P(Y|\text{not }X)

. First, identify what you know: 35% have condition X, so 65% don't have X. Since 12% have both conditions and 35% have X total, then 35% - 12% = 23% have only X. This means 28% - 12% = 16% have only Y. Among the 65% without condition X, 16% have condition Y. Therefore:

P(Y|\text{not }X) = \frac{16\%}{65\%} = \frac{0.16}{0.65} = 0.246

. Answer A (0.123) likely comes from incorrectly using the intersection probability (12%) divided by something inappropriate. Answer B (0.160) might result from confusing the conditional relationship or using 16% directly without proper conditioning. Answer D (0.280) represents the unconditional probability of having condition Y, ignoring the given information that the patient doesn't have condition X. The correct answer is C (0.246). Remember this pattern: conditional probability questions require you to restrict your sample space to only those cases where the condition is met. Always double-check that your denominator reflects the appropriate subset of the population, not the entire original sample. Drawing a 2×2 table or Venn diagram can help visualize these relationships clearly.

Question 18

In a pharmaceutical study, drug effectiveness follows: P(effective) = 0.75, P(side effects) = 0.30, P(effective AND no side effects) = 0.60. Given that a drug causes side effects, what is the probability that it is still effective?

0.15
0.20
0.50 (correct answer)
0.75
0.80

P(A|B) = \frac{P(A \cap B)}{P(B)}

. You're looking for P(effective | side effects). First, identify what you need: P(effective AND side effects) in the numerator, and P(side effects) in the denominator. You're given P(effective AND no side effects) = 0.60, but you need P(effective AND side effects). Since P(effective) = 0.75 total, and 0.60 of that is effective with no side effects, then P(effective AND side effects) = 0.75 - 0.60 = 0.15. Now apply the formula:

P(\text{effective | side effects}) = \frac{0.15}{0.30} = 0.50

Choice A (0.15) represents P(effective AND side effects), not the conditional probability. Choice B (0.20) might result from incorrectly using P(no side effects) = 0.70 in the denominator instead of P(side effects) = 0.30. Choice D (0.75) is simply P(effective), ignoring the conditioning entirely. The correct answer is C (0.50). Study tip: When you see "given that" language, immediately think conditional probability. Always identify what goes in both the numerator (joint probability) and denominator (probability of the given condition) before calculating. Draw a probability tree or contingency table if the relationships aren't immediately clear.

Question 19

In a clinical study, patients can have side effects from medication (20% probability), dietary factors (25% probability), or both (8% probability). What is the probability that a patient experiencing side effects from medication will also experience side effects from dietary factors?

0.080
0.200
0.320
0.400 (correct answer)
0.500

P(A|B) = \frac{P(A \cap B)}{P(B)}

, which reads as "the probability of A given B equals the probability of both A and B divided by the probability of B." Here, you're looking for the probability of dietary side effects given that medication side effects occur. Using our formula:

P(\text{dietary}|\text{medication}) = \frac{P(\text{both})}{P(\text{medication})} = \frac{0.08}{0.20} = 0.40

Answer choice A (0.080) is simply the probability of both events occurring together, not the conditional probability you need. This is a common trap where students confuse joint probability with conditional probability. Answer choice B (0.200) represents the original probability of medication side effects, which isn't what the question asks for. This suggests misreading the question. Answer choice C (0.320) might result from incorrectly trying to subtract probabilities or misapplying probability rules. Some students might calculate

0.20 + 0.25 - 0.08 = 0.37

and round to 0.32, but this gives the probability of experiencing either type of side effect, not the conditional probability. The correct answer is D (0.400). Study tip: Always identify what type of probability you're calculating. Conditional probability questions often use phrases like "given that" or "will also experience." Remember that conditional probability can be greater than either individual probability because you're narrowing your sample space to only cases where the given condition occurs.

Question 20

In a hospital, 45% of patients are male, 60% are over age 65, and 30% are male AND over age 65. If a randomly selected patient is male, what is the probability that the patient is age 65 or younger?

0.150
0.333 (correct answer)
0.400
0.600
0.667

Explanation: This question tests conditional probability, a fundamental concept in biostatistics where you need to find the probability of one event given that another event has already occurred. You're looking for P(age ≤ 65 | male), which reads as "the probability a patient is 65 or younger, given that the patient is male." Using the conditional probability formula: P(A|B) = P(A ∩ B) / P(B). First, identify what you know: P(male) = 0.45, P(over 65) = 0.60, and P(male AND over 65) = 0.30. Since P(age ≤ 65) = 1 - P(over 65), you have P(age ≤ 65) = 1 - 0.60 = 0.40. To find P(male AND age ≤ 65), use: P(male AND age ≤ 65) = P(male) - P(male AND over 65) = 0.45 - 0.30 = 0.15. Therefore: P(age ≤ 65 | male) = P(male AND age ≤ 65) / P(male) = 0.15 / 0.45 = 0.333. Answer A (0.150) is the joint probability P(male AND age ≤ 65), not the conditional probability. Answer C (0.400) is P(age ≤ 65) without conditioning on being male. Answer D (0.600) is P(over 65), which is the opposite age group entirely. Remember: conditional probability narrows your sample space. Once you know the patient is male, you're only considering the 45% of patients who are male, then finding what fraction of those are 65 or younger.