Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games

Opening subject page...

Loading your content

  1. My Subjects
  2. ISEE Middle Level Quantitative Reasoning
  3. Flashcards

ISEE Middle Level Quantitative Reasoning Flashcards: Basic Probability

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

← Back to flashcard decks

What this deck covers

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

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.

ISEE Middle Level Quantitative Reasoning Flashcards: Basic Probability

1

/ 25

0 reviewed

0% Complete

0 reviewing
QUESTION

A bag has 333 red and 222 blue marbles. What is P(red)P(\text{red})P(red) on one draw?

Tap or drag to reveal answer

ANSWER

35\frac{3}{5}53​. There are 3 red marbles out of 5 total marbles in the bag.

Swipe Right = I Know It! 🎉

Swipe Left = Still Learning

All flashcards

Flashcard 1: A bag has 333 red and 222 blue marbles. What is P(red)P(\text{red})P(red) on one draw?

Answer: 35\frac{3}{5}53​. There are 3 red marbles out of 5 total marbles in the bag.

Flashcard 2: A bag has 333 red and 222 blue marbles. Without replacement, what is P(red then red)P(\text{red then red})P(red then red)?

Answer: 310\frac{3}{10}103​. Without replacement, multiply 35\frac{3}{5}53​ by 24\frac{2}{4}42​ for two red draws.

Flashcard 3: A bag has 333 red and 222 blue marbles. With replacement, what is P(red then red)P(\text{red then red})P(red then red)?

Answer: 925\frac{9}{25}259​. With replacement, the probability remains 35\frac{3}{5}53​ for each independent draw, so multiply.

Flashcard 4: What is the probability of getting heads on a fair coin flip?

Answer: 12\frac{1}{2}21​. One favorable outcome (heads) out of two equally likely outcomes on a fair coin.

Flashcard 5: What is the probability of an event EEE if it has fff favorable outcomes out of nnn equally likely outcomes?

Answer: P(E)=fnP(E)=\frac{f}{n}P(E)=nf​. Probability is defined as the ratio of the number of favorable outcomes to the total number of equally likely outcomes.

Flashcard 6: What is the probability of an event that is impossible (has 000 favorable outcomes)?

Answer: 000. An impossible event has no favorable outcomes, resulting in a probability of zero.

Flashcard 7: What is the probability of an event that is certain (always occurs)?

Answer: 111. A certain event includes all possible outcomes, so its probability equals one.

Flashcard 8: What is the relationship between an event EEE and its complement EcE^cEc?

Answer: P(Ec)=1−P(E)P(E^c)=1-P(E)P(Ec)=1−P(E). The probability of the complement is one minus the event's probability, as they together cover all outcomes.

Flashcard 9: What is P(E)P(E)P(E) if P(Ec)=38P(E^c)=\frac{3}{8}P(Ec)=83​?

Answer: 58\frac{5}{8}85​. Subtract the probability of the complement from 1 to find the event's probability.

Flashcard 10: What is P(Ec)P(E^c)P(Ec) if P(E)=0.27P(E)=0.27P(E)=0.27?

Answer: 0.730.730.73. The complement's probability is 1 minus the event's probability.

Flashcard 11: What is the probability of rolling a 444 on a fair six-sided die?

Answer: 16\frac{1}{6}61​. One favorable outcome (rolling a 4) out of six equally likely outcomes on a fair die.

Flashcard 12: What is the probability of rolling an even number on a fair six-sided die?

Answer: 12\frac{1}{2}21​. Three even numbers (2, 4, 6) out of six possible outcomes on a fair die.

Flashcard 13: What is the probability of getting at least one head in two fair coin flips?

Answer: 34\frac{3}{4}43​. Calculated as 1 minus the probability of no heads (both tails), which is 14\frac{1}{4}41​.

Flashcard 14: What is the probability of getting two heads in two fair coin flips?

Answer: 14\frac{1}{4}41​. The probability of heads on each independent flip is 12\frac{1}{2}21​, so multiply for both.

Flashcard 15: What is the probability of drawing a heart from a standard 525252-card deck?

Answer: 14\frac{1}{4}41​. There are 13 hearts out of 52 cards in a standard deck.

Flashcard 16: What is the probability of drawing an ace from a standard 525252-card deck?

Answer: 113\frac{1}{13}131​. There are 4 aces out of 52 cards in a standard deck.

Flashcard 17: What is the probability of drawing a red card from a standard 525252-card deck?

Answer: 12\frac{1}{2}21​. There are 26 red cards (hearts and diamonds) out of 52 in a standard deck.

Flashcard 18: What is the probability of drawing a face card (J, Q, or K) from a 525252-card deck?

Answer: 313\frac{3}{13}133​. There are 12 face cards (3 per suit, 4 suits) out of 52 in a standard deck.

Flashcard 19: What is the addition rule for mutually exclusive events AAA and BBB?

Answer: P(A∪B)=P(A)+P(B)P(A\cup B)=P(A)+P(B)P(A∪B)=P(A)+P(B). For mutually exclusive events, the union's probability is the sum since there is no overlap.

Flashcard 20: If events AAA and BBB are mutually exclusive, what is P(A∩B)P(A\cap B)P(A∩B)?

Answer: 000. Mutually exclusive events cannot occur together, so their intersection has zero probability.

Flashcard 21: If P(A)=15P(A)=\frac{1}{5}P(A)=51​ and P(B)=25P(B)=\frac{2}{5}P(B)=52​ and A,BA,BA,B are mutually exclusive, what is P(A∪B)P(A\cup B)P(A∪B)?

Answer: 35\frac{3}{5}53​. Add the probabilities since the events are mutually exclusive with no overlap.

Flashcard 22: What is the general addition rule for any events AAA and BBB (not necessarily disjoint)?

Answer: P(A∪B)=P(A)+P(B)−P(A∩B)P(A\cup B)=P(A)+P(B)-P(A\cap B)P(A∪B)=P(A)+P(B)−P(A∩B). The general rule accounts for overlap by subtracting the intersection's probability.

Flashcard 23: If P(A)=0.6P(A)=0.6P(A)=0.6, P(B)=0.5P(B)=0.5P(B)=0.5, and P(A∩B)=0.2P(A\cap B)=0.2P(A∩B)=0.2, what is P(A∪B)P(A\cup B)P(A∪B)?

Answer: 0.90.90.9. Apply the general addition rule: add probabilities and subtract the intersection.

Flashcard 24: What is the multiplication rule for independent events AAA and BBB?

Answer: P(A∩B)=P(A)P(B)P(A\cap B)=P(A)P(B)P(A∩B)=P(A)P(B). For independent events, the joint probability is the product of individual probabilities.

Flashcard 25: If P(A)=310P(A)=\frac{3}{10}P(A)=103​ and P(B)=12P(B)=\frac{1}{2}P(B)=21​ and A,BA,BA,B are independent, what is P(A∩B)P(A\cap B)P(A∩B)?

Answer: 320\frac{3}{20}203​. Multiply the probabilities since the events are independent.