# Worksheet: Probability Using Sample Spaces ## Section A: Multiple Choice Questions
Q1: A fair coin is tossed twice. What is the sample space for this experiment? (a) {H, T} (b) {HH, HT, TH, TT} (c) {HH, HT, TT} (d) {H, T, HH, TT}
Solution:
Ans: (b) Explanation: The sample space is the set of all possible outcomes. When a coin is tossed twice, each toss can result in either heads (H) or tails (T). The complete list of outcomes is {HH, HT, TH, TT}, representing all combinations of the two tosses.
Q2: A standard die is rolled once. What is the probability of rolling a number greater than 4? (a) \(\frac{1}{6}\) (b) \(\frac{1}{3}\) (c) \(\frac{1}{2}\) (d) \(\frac{2}{3}\)
Solution:
Ans: (b) Explanation: The sample space for rolling a die is {1, 2, 3, 4, 5, 6}. Numbers greater than 4 are {5, 6}, which gives us 2 favorable outcomes. The probability is \(\frac{2}{6} = \frac{1}{3}\).
Q3: Two dice are rolled simultaneously. How many outcomes are in the sample space? (a) 12 (b) 24 (c) 36 (d) 18
Solution:
Ans: (c) Explanation: Each die has 6 possible outcomes. When rolling two dice, we use the fundamental counting principle: \(6 \times 6 = 36\) total outcomes in the sample space.
Q4: A bag contains 3 red marbles and 2 blue marbles. If one marble is drawn at random, what is the probability of drawing a red marble? (a) \(\frac{2}{5}\) (b) \(\frac{3}{5}\) (c) \(\frac{1}{2}\) (d) \(\frac{3}{2}\)
Solution:
Ans: (b) Explanation: The sample space consists of 5 marbles total (3 red + 2 blue). The number of favorable outcomes is 3 (red marbles). The probability is \(\frac{3}{5}\). Option (d) is impossible since probabilities cannot exceed 1.
Q5: A spinner is divided into 8 equal sections numbered 1 through 8. What is the probability of landing on an even number? (a) \(\frac{1}{4}\) (b) \(\frac{3}{8}\) (c) \(\frac{1}{2}\) (d) \(\frac{5}{8}\)
Solution:
Ans: (c) Explanation: The sample space is {1, 2, 3, 4, 5, 6, 7, 8}. The even numbers are {2, 4, 6, 8}, which gives 4 favorable outcomes. The probability is \(\frac{4}{8} = \frac{1}{2}\).
Q6: A card is drawn from a standard deck of 52 cards. What is the probability of drawing a heart? (a) \(\frac{1}{13}\) (b) \(\frac{1}{4}\) (c) \(\frac{1}{2}\) (d) \(\frac{4}{13}\)
Solution:
Ans: (b) Explanation: A standard deck has 52 cards with 4 suits. Each suit (hearts, diamonds, clubs, spades) has 13 cards. The probability of drawing a heart is \(\frac{13}{52} = \frac{1}{4}\).
Q7: A coin is flipped three times. What is the probability of getting exactly two heads? (a) \(\frac{1}{8}\) (b) \(\frac{1}{4}\) (c) \(\frac{3}{8}\) (d) \(\frac{1}{2}\)
Solution:
Ans: (c) Explanation: The sample space for three coin flips is {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, containing 8 outcomes. Exactly two heads occurs in {HHT, HTH, THH}, which is 3 outcomes. The probability is \(\frac{3}{8}\).
Q8: An experiment has a sample space S = {A, B, C, D, E}. If each outcome is equally likely, what is the probability of event {A, C, E}? (a) \(\frac{1}{5}\) (b) \(\frac{2}{5}\) (c) \(\frac{3}{5}\) (d) \(\frac{4}{5}\)
Solution:
Ans: (c) Explanation: The sample space has 5 equally likely outcomes. The event {A, C, E} contains 3 favorable outcomes. Since outcomes are equally likely, the probability is \(\frac{3}{5}\).
## Section B: Fill in the Blanks Q9: The set of all possible outcomes of an experiment is called the __________.
Solution:
Ans: sample space Explanation: The sample space is the fundamental concept in probability that lists every possible outcome of a random experiment.
Q10: If each outcome in a sample space has the same chance of occurring, the outcomes are said to be __________.
Solution:
Ans: equally likely Explanation: When all outcomes have the same probability of occurring, they are called equally likely outcomes. This is an important assumption in many probability problems.
Q11: The probability of any event ranges from __________ to __________ (inclusive).
Solution:
Ans: 0 to 1 (or 0, 1) Explanation: The probability of an event must be between 0 and 1, where 0 represents an impossible event and 1 represents a certain event.
Q12: If a die is rolled, the probability of getting a 7 is __________ because this outcome is not in the sample space.
Solution:
Ans: 0 Explanation: Since a standard die only has faces numbered 1 through 6, rolling a 7 is impossible. An impossible event has a probability of 0.
Q13: The number of favorable outcomes divided by the total number of outcomes in the sample space gives the __________ of an event.
Solution:
Ans: probability Explanation: This is the classical definition of probability: \(P(E) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}\).
Q14: When flipping a coin and rolling a die simultaneously, the total number of outcomes in the sample space is __________.
Solution:
Ans: 12 Explanation: Using the fundamental counting principle, a coin has 2 outcomes and a die has 6 outcomes, so the total is \(2 \times 6 = 12\) outcomes.
## Section C: Word Problems Q15: A jar contains 5 green marbles, 3 yellow marbles, and 2 orange marbles. If one marble is selected at random, what is the probability that it is either green or yellow?
Solution:
Ans: Total marbles = 5 + 3 + 2 = 10 Favorable outcomes (green or yellow) = 5 + 3 = 8 Probability = \(\frac{8}{10} = \frac{4}{5}\) Final Answer: \(\frac{4}{5}\) or 0.8
Q16: Emma spins a spinner that is divided into 12 equal sections numbered 1 through 12. What is the probability that the spinner lands on a multiple of 3?
Solution:
Ans: Sample space = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} Multiples of 3 = {3, 6, 9, 12} Number of favorable outcomes = 4 Total outcomes = 12 Probability = \(\frac{4}{12} = \frac{1}{3}\) Final Answer: \(\frac{1}{3}\)
Q17: Two coins are tossed simultaneously. Find the probability of getting at least one head.
Solution:
Ans: Sample space = {HH, HT, TH, TT} Total outcomes = 4 Outcomes with at least one head = {HH, HT, TH} Favorable outcomes = 3 Probability = \(\frac{3}{4}\) Final Answer: \(\frac{3}{4}\) or 0.75
Q18: A number is randomly selected from the set {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}. What is the probability that the number selected is a prime number?
Solution:
Ans: Sample space = {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} Total outcomes = 11 Prime numbers in the set = {11, 13, 17, 19} Favorable outcomes = 4 Probability = \(\frac{4}{11}\) Final Answer: \(\frac{4}{11}\)
Q19: A class has 15 boys and 12 girls. If one student is randomly selected to be the class representative, what is the probability that a girl is selected?
Solution:
Ans: Total students = 15 + 12 = 27 Number of girls = 12 Probability = \(\frac{12}{27} = \frac{4}{9}\) Final Answer: \(\frac{4}{9}\)
Q20: A letter is randomly chosen from the word "PROBABILITY". What is the probability that the letter chosen is a vowel?
Solution:
Ans: The word "PROBABILITY" has 11 letters: P-R-O-B-A-B-I-L-I-T-Y Vowels in the word: O, A, I, I (4 vowels) Total letters = 11 Favorable outcomes = 4 Probability = \(\frac{4}{11}\) Final Answer: \(\frac{4}{11}\)
The document Worksheet (with Solutions): Probability Using Sample Spaces is a part of the Grade 9 Course Statistics & Probability.
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