PSAT Math Flashcards: Probability

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

PSAT Math

Probability

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QUESTION
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What condition must hold for events AA and BB to be independent using conditional probability?

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ANSWER

P(AB)=P(A)P(A\mid B)=P(A). Independence means BB doesn't change AA's probability.

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

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Flashcard 1: What condition must hold for events AA and BB to be independent using conditional probability?

Answer: P(AB)=P(A)P(A\mid B)=P(A). Independence means BB doesn't change AA's probability.

Flashcard 2: What is the formula for the expected value of a discrete variable with outcomes xix_i and probabilities pip_i?

Answer: E=xipiE=\sum x_i p_i. Sum of each outcome times its probability.

Flashcard 3: A bag has 33 red and 22 blue marbles. What is P(red)P(\text{red}) in one draw?

Answer: 35\frac{3}{5}. Three red marbles out of five total marbles.

Flashcard 4: What is the complement rule for an event AA?

Answer: P(Ac)=1P(A)P(A^c)=1-P(A). The probability of not-AA equals one minus the probability of AA.

Flashcard 5: What is the definition 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)}. Ratio of joint probability to the condition's probability.

Flashcard 6: Find P(AB)P(A\cap B) if AA and BB are independent, P(A)=0.3P(A)=0.3, and P(B)=0.4P(B)=0.4.

Answer: 0.120.12. For independent events: P(AB)=0.3×0.4=0.12P(A \cap B) = 0.3 \times 0.4 = 0.12.

Flashcard 7: What is the probability of a union if P(A)=0.3P(A)=0.3, P(B)=0.5P(B)=0.5, and P(AB)=0.1P(A\cap B)=0.1?

Answer: P(AB)=0.7P(A\cup B)=0.7. Apply general addition rule: 0.3+0.50.1=0.70.3+0.5-0.1=0.7.

Flashcard 8: What is P(AB)P(A\cup B) if P(A)=0.5P(A)=0.5, P(B)=0.6P(B)=0.6, and P(AB)=0.2P(A\cap B)=0.2?

Answer: 0.90.9. Apply general addition: 0.5+0.60.2=0.90.5 + 0.6 - 0.2 = 0.9.

Flashcard 9: Find P(AB)P(A\cup B) if AA and BB are disjoint, P(A)=0.25P(A)=0.25, and P(B)=0.40P(B)=0.40.

Answer: 0.650.65. Disjoint events: simply add 0.25+0.40=0.650.25+0.40=0.65.

Flashcard 10: What is the formula connecting conditional probability to intersection: P(AB)P(A\cap B)?

Answer: P(AB)=P(AB)P(B)P(A\cap B)=P(A\mid B)\cdot P(B). Rearranges conditional probability to find intersection.

Flashcard 11: What is the formula for probability of event AA using favorable and total outcomes?

Answer: P(A)=favorabletotalP(A)=\frac{\text{favorable}}{\text{total}}. Divide favorable outcomes by total possible outcomes.

Flashcard 12: What is the expected value of a game paying $5 with probability 0.20.2 and $0 otherwise?

Answer: 11. Expected value: 5×0.2+0×0.8=15\times 0.2+0\times 0.8=1.

Flashcard 13: What is the probability of flipping exactly one head in two fair coin flips?

Answer: 12\frac{1}{2}. Outcomes HT and TH out of HH, HT, TH, TT.

Flashcard 14: What is the probability range for any event EE?

Answer: 0P(E)10\le P(E)\le 1. Probability is always between impossible (0) and certain (1).

Flashcard 15: Identify whether AA and BB are independent if 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: Independent. Check: P(A)×P(B)=0.4×0.5=0.2=P(AB)P(A) \times P(B) = 0.4 \times 0.5 = 0.2 = P(A \cap B).

Flashcard 16: What is the addition rule for disjoint events AA and BB?

Answer: If AB=A\cap B=\varnothing, then P(AB)=P(A)+P(B)P(A\cup B)=P(A)+P(B). Disjoint events have no overlap, so just add their probabilities.

Flashcard 17: A bag has 55 red and 33 blue marbles. One marble is drawn. What is P(red)P(\text{red})?

Answer: 58\frac{5}{8}. Five red marbles out of eight total marbles.

Flashcard 18: What is the range of possible values for any probability P(A)P(A)?

Answer: 0P(A)10\le P(A)\le 1. Probabilities must be between 0 and 1 inclusive.

Flashcard 19: What is P(not 5)P(\text{not }5) when rolling one fair six-sided die?

Answer: 56\frac{5}{6}. Five favorable outcomes (1,2,3,4,6) out of six.

Flashcard 20: A bag has 55 red and 33 blue marbles. What is P(red)P(\text{red}) on one draw?

Answer: 58\frac{5}{8}. 5 red marbles out of 8 total marbles.

Flashcard 21: A bag has 55 red and 33 blue marbles. Two are drawn with replacement. What is P(both red)P(\text{both red})?

Answer: (58)2=2564\left(\frac{5}{8}\right)^2=\frac{25}{64}. With replacement, each draw has 58\frac{5}{8} probability; multiply for both.

Flashcard 22: What is the formula for the complement rule for an event AA?

Answer: P(Ac)=1P(A)P(A^c)=1-P(A). The probability of not-AA equals one minus the probability of AA.

Flashcard 23: A fair die is rolled once. What is P(roll is a multiple of 3)P(\text{roll is a multiple of }3)?

Answer: 13\frac{1}{3}. Multiples of 3 are {3,6}, so 26=13\frac{2}{6}=\frac{1}{3}.

Flashcard 24: Find P(AB)P(A\cup B) if P(A)=0.5P(A)=0.5, P(B)=0.3P(B)=0.3, and P(AB)=0.1P(A\cap B)=0.1.

Answer: 0.70.7. Use addition rule: 0.5+0.30.1=0.70.5+0.3-0.1=0.7.

Flashcard 25: Find P(AB)P(A\cup B) if P(A)=0.6P(A)=0.6, P(B)=0.5P(B)=0.5, and P(AB)=0.2P(A\cap B)=0.2.

Answer: 0.90.9. Apply general addition rule: 0.6+0.50.2=0.90.6 + 0.5 - 0.2 = 0.9.

Flashcard 26: What is P(AB)P(A\cap B) if AA and BB are independent, P(A)=0.3P(A)=0.3, and P(B)=0.6P(B)=0.6?

Answer: 0.180.18. Independent events: 0.3×0.6=0.180.3\times 0.6=0.18.

Flashcard 27: A bag has 33 red and 22 blue marbles. With replacement, what is P(red then blue)P(\text{red then blue})?

Answer: 625\frac{6}{25}. 35×25\frac{3}{5}\times\frac{2}{5} since marble is replaced.

Flashcard 28: A bag has 33 red and 22 blue marbles. Two are drawn without replacement. What is P(both red)P(\text{both red})?

Answer: 310\frac{3}{10}. First red 35\frac{3}{5}, then second red 24\frac{2}{4}: 35×24=310\frac{3}{5}\times\frac{2}{4}=\frac{3}{10}.

Flashcard 29: What does it mean for events AA and BB to be mutually exclusive?

Answer: P(AB)=0P(A\cap B)=0. Mutually exclusive events cannot occur simultaneously.

Flashcard 30: Find P(AB)P(A\mid B) if P(AB)=0.15P(A\cap B)=0.15 and P(B)=0.5P(B)=0.5.

Answer: 0.30.3. Apply conditional probability: P(AB)=0.150.5=0.3P(A|B) = \frac{0.15}{0.5} = 0.3.

Flashcard 31: If P(A)=0.3P(A)=0.3, P(B)=0.5P(B)=0.5, and AA and BB are independent, what is P(AB)P(A\cap B)?

Answer: 0.150.15. For independent events, P(AB)=P(A)P(B)=0.3×0.5P(A\cap B)=P(A)\cdot P(B)=0.3\times 0.5.

Flashcard 32: What condition defines independence using probabilities of AA and BB?

Answer: A,BA,B independent if P(AB)=P(A)P(B)P(A\cap B)=P(A)P(B). Events are independent when their joint probability equals the product.

Flashcard 33: What is the formula for conditional probability P(AB)P(A\mid B) (with P(B)>0P(B)>0)?

Answer: P(AB)=P(AB)P(B)P(A\mid B)=\frac{P(A\cap B)}{P(B)}. Divide the joint probability by the condition's probability.

Flashcard 34: What is P(AB)P(A\cap B) if P(A)=0.3P(A)=0.3 and P(B)=0.4P(B)=0.4 and the events are independent?

Answer: 0.120.12. Independent events: 0.3×0.4=0.120.3\times 0.4=0.12.

Flashcard 35: What is P(at least one head)P(\text{at least one head}) when flipping 22 fair coins?

Answer: 34\frac{3}{4}. Three favorable (HH, HT, TH) out of four outcomes.

Flashcard 36: What is the conditional probability formula P(AB)P(A\mid B)?

Answer: P(AB)=P(AB)P(B)P(A\mid B)=\frac{P(A\cap B)}{P(B)}. Divides joint probability by the condition's probability.

Flashcard 37: Identify the event described by "AA and BB" using set notation.

Answer: ABA\cap B. Intersection symbol represents "and" in probability.

Flashcard 38: What is P(AB)P(A\cap B) if AA and BB are independent, P(A)=0.3P(A)=0.3, and P(B)=0.4P(B)=0.4?

Answer: 0.120.12. Independent events: P(AB)=0.3×0.4=0.12P(A\cap B) = 0.3 \times 0.4 = 0.12.

Flashcard 39: A fair die is rolled. What is the probability of rolling a number greater than 44?

Answer: 13\frac{1}{3}. Favorable outcomes are 5 and 6, so 26=13\frac{2}{6}=\frac{1}{3}.

Flashcard 40: If P(A)=0.35P(A)=0.35, what is P(Ac)P(A^c)?

Answer: 0.650.65. Using complement rule: 10.35=0.651-0.35=0.65

Flashcard 41: A bag has 33 red and 22 blue marbles. Two are drawn without replacement. What is P(both red)P(\text{both red})?

Answer: 310\frac{3}{10}. First red 35\frac{3}{5}, then 24\frac{2}{4} remaining red: 35×24=310\frac{3}{5}\times\frac{2}{4}=\frac{3}{10}.

Flashcard 42: What is the multiplication rule for any events AA and BB using conditional probability?

Answer: P(AB)=P(B)P(AB)P(A\cap B)=P(B)P(A\mid B). General multiplication rule using conditional probability.

Flashcard 43: What is the probability of drawing two aces without replacement from a 5252-card deck?

Answer: 1221\frac{1}{221}. 452×351=122652=1221\frac{4}{52}\times\frac{3}{51}=\frac{12}{2652}=\frac{1}{221}.

Flashcard 44: What is the conditional probability formula for P(AB)P(A\mid B)?

Answer: P(AB)=P(AB)P(B)P(A\mid B)=\frac{P(A\cap B)}{P(B)}. Probability of A given B equals their intersection over B.

Flashcard 45: Find 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. Use multiplication rule: 0.3×0.5=0.150.3\times 0.5=0.15.

Flashcard 46: What is the fundamental counting principle for mm choices then nn choices?

Answer: mnm\cdot n total outcomes. Multiply the number of choices at each step.

Flashcard 47: What is P(AB)P(A\mid B) if P(AB)=0.12P(A\cap B)=0.12 and P(B)=0.30P(B)=0.30?

Answer: 0.400.40. Divide joint probability by condition: 0.120.30=0.40\frac{0.12}{0.30}=0.40.

Flashcard 48: Which option equals 5C2^5C_2 (combinations of 55 items chosen 22)?

Answer: 1010. 5C2=5!2!3!=202=10^5C_2 = \frac{5!}{2!3!} = \frac{20}{2} = 10 ways to choose 2 from 5.

Flashcard 49: What is the probability that a fair die roll is even?

Answer: 12\frac{1}{2}. Three even outcomes (2,4,6) out of six total outcomes.

Flashcard 50: What is the multiplication rule using conditional probability for events AA and BB?

Answer: P(AB)=P(AB)P(B)P(A\cap B)=P(A\mid B)\cdot P(B). Rearranges conditional probability definition to find intersection probability.

Flashcard 51: What is the probability of exactly kk successes in nn independent trials with success rate pp?

Answer: (nk)pk(1p)nk\binom{n}{k}p^k(1-p)^{n-k}. Binomial probability formula for k successes in n trials.

Flashcard 52: A bag has 33 red and 22 blue marbles. Without replacement, what is P(red then blue)P(\text{red then blue})?

Answer: 310\frac{3}{10}. P(red)×P(bluered)=35×24=310P(\text{red})\times P(\text{blue}|\text{red}) = \frac{3}{5}\times\frac{2}{4} = \frac{3}{10}.

Flashcard 53: What is the probability that a fair die roll is greater than 44?

Answer: 13\frac{1}{3}. Two favorable outcomes (5,6) out of six total outcomes.

Flashcard 54: What is the probability rule for "at least one" occurrence of event AA in nn trials?

Answer: P(1)=1P(0)P(\ge 1)=1-P(0). At least one equals 1 minus the probability of none.

Flashcard 55: Identify whether AA and BB are mutually exclusive if P(A)=0.6P(A)=0.6, P(B)=0.5P(B)=0.5, and P(AB)=0.2P(A\cap B)=0.2.

Answer: Not mutually exclusive. Since P(AB)=0.20P(A\cap B)=0.2\neq 0, events can occur together.

Flashcard 56: Find the probability of at least one head in 22 fair coin flips.

Answer: 34\frac{3}{4}. All outcomes except TT: 34\frac{3}{4} (HH, HT, TH).

Flashcard 57: What is the formula for permutations of nn distinct items taken rr at a time?

Answer: n!(nr)!\frac{n!}{(n-r)!}. Order matters; divide n!n! by (nr)!(n-r)! to remove unused items.

Flashcard 58: A fair coin is flipped twice. What is P(exactly one head)P(\text{exactly one head})?

Answer: 12\frac{1}{2}. Outcomes HT and TH out of four equally likely: 24=12\frac{2}{4} = \frac{1}{2}.

Flashcard 59: Find P(Ac)P(A^c) if P(A)=0.37P(A)=0.37.

Answer: 0.630.63. Apply complement rule: P(Ac)=10.37=0.63P(A^c) = 1 - 0.37 = 0.63.

Flashcard 60: A fair die is rolled once. What is P(prime)P(\text{prime})?

Answer: 12\frac{1}{2}. Prime numbers on a die are 22, 33, and 55: three out of six outcomes.

Flashcard 61: What is the formula for probability of an event AA using favorable and total outcomes?

Answer: P(A)=favorable outcomestotal outcomesP(A)=\frac{\text{favorable outcomes}}{\text{total outcomes}}. Ratio of successful outcomes to all possible outcomes.

Flashcard 62: What is the formula for the complement of an event AA?

Answer: P(Ac)=1P(A)P(A^c)=1-P(A). The complement has probability one minus the original.

Flashcard 63: If P(AB)=0.12P(A\cap B)=0.12 and P(B)=0.3P(B)=0.3, what is P(AB)P(A\mid B)?

Answer: 0.40.4. Apply conditional probability formula: 0.120.3=0.4\frac{0.12}{0.3} = 0.4.

Flashcard 64: A bag has 33 red and 22 blue marbles. With replacement, what is P(red then blue)P(\text{red then blue})?

Answer: 625\frac{6}{25}. With replacement: P(red)P(blue)=3525=625P(\text{red})\cdot P(\text{blue})=\frac{3}{5}\cdot\frac{2}{5}=\frac{6}{25}

Flashcard 65: Find P(AB)P(A\cap B) if P(A)=0.60P(A)=0.60 and P(BA)=0.20P(B\mid A)=0.20.

Answer: 0.120.12. Apply multiplication rule: 0.60×0.20=0.120.60\times 0.20=0.12.

Flashcard 66: Find P(AB)P(A\cup B) if P(A)=0.5P(A)=0.5, P(B)=0.4P(B)=0.4, and P(AB)=0.2P(A\cap B)=0.2.

Answer: 0.70.7. Apply union formula: 0.5+0.40.2=0.70.5+0.4-0.2=0.7.

Flashcard 67: What is the counting formula for permutations of nn items taken rr at a time?

Answer: nPr=n!(nr)!^nP_r=\frac{n!}{(n-r)!}. Order matters: arrange r items from n total items.

Flashcard 68: What is the multiplication rule using conditional probability for events AA and BB?

Answer: P(AB)=P(B)P(AB)P(A\cap B)=P(B)\cdot P(A\mid B). Rearranges conditional probability formula to find intersection.

Flashcard 69: What condition must hold for events AA and BB to be mutually exclusive (disjoint)?

Answer: P(AB)=0P(A\cap B)=0. Disjoint events cannot occur simultaneously.

Flashcard 70: How many ways can you choose 22 students from 55 students?

Answer: (52)=10\binom{5}{2}=10. Combinations formula: 5!2!3!=10\frac{5!}{2!3!}=10.

Flashcard 71: What is the probability of the empty event \varnothing?

Answer: P()=0P(\varnothing)=0. The impossible event has probability 0.

Flashcard 72: If P(A)=0.6P(A)=0.6, P(B)=0.5P(B)=0.5, and P(AB)=0.2P(A\cap B)=0.2, what is P(AB)P(A\cup B)?

Answer: 0.90.9. Apply addition rule: 0.6+0.50.2=0.90.6+0.5-0.2=0.9.

Flashcard 73: What is the definition of probability for an event with equally likely outcomes?

Answer: P(E)=favorable outcomestotal outcomesP(E)=\frac{\text{favorable outcomes}}{\text{total outcomes}}. Ratio of successful outcomes to all possible outcomes.

Flashcard 74: What is the complement rule for an event EE?

Answer: P(Ec)=1P(E)P(E^c)=1-P(E). The complement's probability is one minus the event's probability.

Flashcard 75: What is the counting formula for combinations of nn distinct items chosen rr at a time?

Answer: n!r!(nr)!\frac{n!}{r!(n-r)!}. Order doesn't matter; divide by r!r! to remove arrangements.

Flashcard 76: A bag has 33 red and 22 blue marbles. What is P(red)P(\text{red}) on one draw?

Answer: 35\frac{3}{5}. Three red marbles out of five total marbles.

Flashcard 77: If P(A)=0.6P(A)=0.6, P(B)=0.5P(B)=0.5, and AA and BB are independent, what is P(AB)P(A\cap B)?

Answer: 0.30.3. For independent events: 0.6×0.5=0.30.6\times 0.5=0.3

Flashcard 78: What is the key difference: independent vs. mutually exclusive events (in one statement)?

Answer: Independent: P(AB)=P(A)P(B)P(A\cap B)=P(A)P(B); disjoint: P(AB)=0P(A\cap B)=0. Independent can overlap; disjoint cannot occur together.

Flashcard 79: What is the definition of probability of event AA using outcomes and favorable outcomes?

Answer: P(A)=favorable outcomestotal outcomesP(A)=\frac{\text{favorable outcomes}}{\text{total outcomes}}. Ratio of successful outcomes to all possible outcomes.

Flashcard 80: What is the probability of rolling an even number on a fair six-sided die?

Answer: 12\frac{1}{2}. Three even outcomes (2,4,6) out of six total outcomes.

Flashcard 81: A bag has 33 red and 22 blue marbles. Two are drawn with replacement. What is P(both red)P(\text{both red})?

Answer: 925\frac{9}{25}. With replacement: 35×35=925\frac{3}{5}\times\frac{3}{5}=\frac{9}{25}.

Flashcard 82: What is the formula for the probability of ABA\cup B for any two events AA and BB?

Answer: P(AB)=P(A)+P(B)P(AB)P(A\cup B)=P(A)+P(B)-P(A\cap B). Add individual probabilities, subtract overlap to avoid double-counting.

Flashcard 83: What condition must hold for events AA and BB to be mutually exclusive (disjoint)?

Answer: P(AB)=0P(A\cap B)=0. Mutually exclusive events cannot occur together.

Flashcard 84: A fair die is rolled once. What is P(prime)P(\text{prime}) where primes are 2,3,52,3,5?

Answer: 12\frac{1}{2}. Three primes (2,3,5)(2,3,5) out of six outcomes: 36=12\frac{3}{6}=\frac{1}{2}.

Flashcard 85: What is the probability of not rolling a 66 on one fair die roll?

Answer: 56\frac{5}{6}. Five favorable outcomes out of six total.

Flashcard 86: What condition must hold for events AA and BB to be independent?

Answer: P(AB)=P(A)P(A\mid B)=P(A). Independence means conditioning doesn't change the probability.

Flashcard 87: What is the definition of 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)}. Probability of A given B equals their intersection divided by B's probability.

Flashcard 88: What is P(Ac)P(A^c) if P(A)=0.37P(A)=0.37?

Answer: 0.630.63. Apply complement rule: 10.37=0.631-0.37=0.63.

Flashcard 89: What condition defines mutual exclusivity (disjointness) for events AA and BB?

Answer: Mutually exclusive if P(AB)=0P(A\cap B)=0. Events cannot occur together when their intersection is empty.

Flashcard 90: A card is drawn from a standard 5252-card deck. What is P(heart)P(\text{heart})?

Answer: 14\frac{1}{4}. There are 1313 hearts in a standard 5252-card deck.

Flashcard 91: What is the complement rule for an event AA in probability notation?

Answer: P(Ac)=1P(A)P(A^c)=1-P(A). The probability of not-A equals 1 minus the probability of A.

Flashcard 92: What is the condition for events AA and BB to be mutually exclusive?

Answer: P(AB)=0P(A\cap B)=0. Mutually exclusive events cannot happen simultaneously.

Flashcard 93: Identify whether AA and BB are independent if P(A)=0.5P(A)=0.5, P(B)=0.2P(B)=0.2, and P(AB)=0.1P(A\cap B)=0.1.

Answer: Independent. Check if P(AB)=P(A)P(B)P(A\cap B)=P(A)P(B): 0.1=0.5×0.20.1=0.5×0.2 ✓.

Flashcard 94: If events AA and BB are mutually exclusive with P(A)=0.3P(A)=0.3 and P(B)=0.5P(B)=0.5, what is P(AB)P(A\cup B)?

Answer: 0.80.8. For mutually exclusive events, add probabilities: 0.3+0.5=0.80.3 + 0.5 = 0.8.

Flashcard 95: What is the expected value formula for a discrete random variable XX?

Answer: E(X)=xP(X=x)E(X)=\sum x\,P(X=x). Sum each value times its probability for discrete variables.

Flashcard 96: Find the probability of exactly one head in 22 fair coin flips.

Answer: 12\frac{1}{2}. Outcomes HT and TH out of 4 total: 24=12\frac{2}{4} = \frac{1}{2}.

Flashcard 97: Which condition must hold for events AA and BB to be independent (in probability form)?

Answer: P(AB)=P(A)P(A\mid B)=P(A). Independent events satisfy: knowing BB occurred doesn't change P(A)P(A).

Flashcard 98: A fair coin is flipped 33 times. What is the probability of exactly 22 heads?

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

Flashcard 99: Two fair dice are rolled. What is P(sum=7)P(\text{sum}=7)?

Answer: 16\frac{1}{6}. Six ways to sum to 77: (1,6)(1,6), (2,5)(2,5), (3,4)(3,4), (4,3)(4,3), (5,2)(5,2), (6,1)(6,1) out of 3636.

Flashcard 100: Find P(AB)P(A\mid B) if P(AB)=0.12P(A\cap B)=0.12 and P(B)=0.3P(B)=0.3.

Answer: 0.40.4. Divide intersection by condition: 0.120.3=0.4\frac{0.12}{0.3}=0.4.