GRE Quantitative Flashcards: Probability

Study Probability in GRE Quantitative with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

GRE Quantitative

Probability

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QUESTION
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Identify P(AB)P(A\cap B) if P(A)=0.3P(A)=0.3 and P(BA)=0.5P(B\mid A)=0.5.

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ANSWER

0.150.15. Apply the multiplication rule: P(AB)=0.3×0.5=0.15P(A\cap B) = 0.3 \times 0.5 = 0.15.

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What this deck covers

This deck focuses on Probability, giving you a quick way to review the definitions, rules, and examples that matter most for GRE Quantitative.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

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Flashcard 1: Identify P(AB)P(A\cap B) if P(A)=0.3P(A)=0.3 and P(BA)=0.5P(B\mid A)=0.5.

Answer: 0.150.15. Apply the multiplication rule: P(AB)=0.3×0.5=0.15P(A\cap B) = 0.3 \times 0.5 = 0.15.

Flashcard 2: What is the definition of probability for an event EE in a finite equally likely sample space?

Answer: P(E)=#(E)#(S)P(E)=\frac{\#(E)}{\#(S)}. In a finite sample space where all outcomes are equally likely, the probability of event EE is the number of favorable outcomes divided by the total number of outcomes.

Flashcard 3: What is the formula for probability of "at least one" occurrence of event AA in repeated trials?

Answer: P(1)=1P(0)P(\ge 1)=1-P(0). The probability of at least one occurrence is one minus the probability of zero occurrences in independent trials.

Flashcard 4: Find P(Ac)P(A^c) given P(A)=0.13P(A)=0.13.

Answer: 0.870.87. Use the complement rule: P(Ac)=10.13=0.87P(A^c) = 1 - 0.13 = 0.87.

Flashcard 5: What is the multiplication rule using conditional probability for events AA and BB?

Answer: P(AB)=P(A)P(BA)P(A\cap B)=P(A)P(B\mid A). The multiplication rule expresses the joint probability as the product of one event's probability and the conditional probability of the other given the first.

Flashcard 6: What is the counting formula for combinations of nn distinct items chosen rr at a time?

Answer: (nr)=n!r!(nr)!\binom{n}{r}=\frac{n!}{r!(n-r)!}. Combinations count the number of ways to choose rr items out of nn distinct ones, where order does not matter.

Flashcard 7: What is the probability of drawing 22 red cards in 22 draws with replacement from a 5252-card deck?

Answer: 14\frac{1}{4}. With replacement, P(P(both red)=2652×2652=12×12=14) = \frac{26}{52} \times \frac{26}{52} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}.

Flashcard 8: What is the counting formula for permutations of nn distinct items taken rr at a time?

Answer: P(n,r)=n!(nr)!P(n,r)=\frac{n!}{(n-r)!}. Permutations count the number of ways to arrange rr items out of nn distinct ones, where order matters.

Flashcard 9: Identify P(AB)P(A\mid B) given P(AB)=0.12P(A\cap B)=0.12 and P(B)=0.3P(B)=0.3.

Answer: 0.40.4. Use the conditional probability formula: P(AB)=0.120.3=0.4P(A\mid B) = \frac{0.12}{0.3} = 0.4.

Flashcard 10: What is the addition rule for two events AA and BB (not necessarily disjoint)?

Answer: P(AB)=P(A)+P(B)P(AB)P(A\cup B)=P(A)+P(B)-P(A\cap B). The addition rule accounts for the overlap by subtracting the intersection probability from the sum of individual probabilities.

Flashcard 11: Identify P(AB)P(A\cup B) given P(A)=0.4P(A)=0.4, P(B)=0.5P(B)=0.5, and P(AB)=0.2P(A\cap B)=0.2.

Answer: 0.70.7. Apply the addition rule: P(AB)=0.4+0.50.2=0.7P(A\cup B) = 0.4 + 0.5 - 0.2 = 0.7.

Flashcard 12: Identify E[X]E[X] for XBin(10,0.3)X\sim\text{Bin}(10,0.3).

Answer: 33. For binomial, E[X]=np=10×0.3=3E[X] = np = 10 \times 0.3 = 3.

Flashcard 13: What is the formula for the variance of XBin(n,p)X\sim\text{Bin}(n,p)?

Answer: Var(X)=np(1p)\text{Var}(X)=np(1-p). The variance of a binomial random variable measures spread as npnp times the failure probability.

Flashcard 14: What is the probability of at least one head in 33 fair coin flips?

Answer: 78\frac{7}{8}. Calculate as 1P(1 - P(all tails)=1(12)3=118=78) = 1 - (\frac{1}{2})^3 = 1 - \frac{1}{8} = \frac{7}{8}.

Flashcard 15: What is the probability of getting at least one 66 in 22 fair die rolls?

Answer: 1136\frac{11}{36}. Calculate as 1P(1 - P(no 6)=1(56)2=12536=1136) = 1 - (\frac{5}{6})^2 = 1 - \frac{25}{36} = \frac{11}{36}.

Flashcard 16: What is the formula for conditional probability P(AB)P(A\mid B)?

Answer: P(AB)=P(AB)P(B)P(A\mid B)=\frac{P(A\cap B)}{P(B)}. Conditional probability measures the likelihood of AA occurring given BB has occurred, by dividing the joint probability by P(B)P(B).

Flashcard 17: Identify P(AB)P(A\cap B) if AA and BB are independent with P(A)=0.2P(A)=0.2 and P(B)=0.6P(B)=0.6.

Answer: 0.120.12. For independent events, P(AB)=0.2×0.6=0.12P(A\cap B) = 0.2 \times 0.6 = 0.12.

Flashcard 18: What is the addition rule for mutually exclusive events AA and BB?

Answer: P(AB)=P(A)+P(B)P(A\cup B)=P(A)+P(B). For mutually exclusive events, there is no overlap, so their union probability is simply the sum of their individual probabilities.

Flashcard 19: What is the formula for the expected value of a binomial random variable XBin(n,p)X\sim\text{Bin}(n,p)?

Answer: E[X]=npE[X]=np. The expected value of a binomial random variable is the product of the number of trials and the success probability per trial.

Flashcard 20: What condition defines independence of events AA and BB using an equation?

Answer: P(AB)=P(A)P(B)P(A\cap B)=P(A)P(B). Events AA and BB are independent if their joint probability equals the product of their marginal probabilities.

Flashcard 21: What is P(AB)P(A\mid B) if AA and BB are independent and P(A)P(A) is known?

Answer: P(AB)=P(A)P(A\mid B)=P(A). For independent events, the occurrence of BB does not affect the probability of AA, so the conditional equals the unconditional probability.

Flashcard 22: What is the formula for the probability of exactly kk successes in nn Bernoulli trials?

Answer: (nk)pk(1p)nk\binom{n}{k}p^k(1-p)^{n-k}. The binomial probability formula gives the likelihood of exactly kk successes in nn independent trials each with success probability pp.

Flashcard 23: What is the probability of exactly 22 heads in 33 fair coin flips?

Answer: 38\frac{3}{8}. Use binomial formula: (32)(12)3=3×18=38\binom{3}{2} (\frac{1}{2})^3 = 3 \times \frac{1}{8} = \frac{3}{8}.

Flashcard 24: What is the complement rule for an event AA?

Answer: P(Ac)=1P(A)P(A^c)=1-P(A). The complement rule states that the probability of an event not occurring equals one minus the probability of it occurring.