# I'll generate a Grade 9 Statistics & Probability worksheet on Dependent and Independent Events
Section A: Multiple Choice Questions
Q1: A fair coin is flipped twice. What is the probability of getting heads on the second flip, given that the first flip resulted in tails? (a) \(\frac{1}{4}\) (b) \(\frac{1}{2}\) (c) \(\frac{3}{4}\) (d) 1
Solution:
Ans: (b) Explanation: The coin flips are independent events. The outcome of the first flip does not affect the probability of the second flip. Since it is a fair coin, the probability of getting heads on any single flip is \(\frac{1}{2}\), regardless of previous outcomes.
Q2: A bag contains 5 red marbles and 3 blue marbles. If two marbles are drawn without replacement, what type of events are the two draws? (a) Independent events (b) Dependent events (c) Mutually exclusive events (d) Complementary events
Solution:
Ans: (b) Explanation: These are dependent events because the first draw changes the composition of the bag, which affects the probability of the second draw. When sampling without replacement, the events are dependent.
Q3: Two dice are rolled simultaneously. What is the probability of rolling a 4 on the first die and a 6 on the second die? (a) \(\frac{1}{36}\) (b) \(\frac{1}{12}\) (c) \(\frac{1}{6}\) (d) \(\frac{1}{3}\)
Solution:
Ans: (a) Explanation: Rolling two dice represents independent events. The probability of rolling a 4 on the first die is \(\frac{1}{6}\) and the probability of rolling a 6 on the second die is \(\frac{1}{6}\). For independent events, we multiply: \(P(A \text{ and } B) = P(A) \times P(B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\).
Q4: A deck of 52 cards has one card drawn and set aside (not replaced). Then a second card is drawn. What is the probability that both cards are aces? (a) \(\frac{1}{169}\) (b) \(\frac{4}{663}\) (c) \(\frac{1}{221}\) (d) \(\frac{1}{13}\)
Solution:
Ans: (c) Explanation: These are dependent events. The probability of the first card being an ace is \(\frac{4}{52} = \frac{1}{13}\). After removing one ace, there are 3 aces left out of 51 cards, so the probability of the second card being an ace is \(\frac{3}{51} = \frac{1}{17}\). Multiplying: \(\frac{1}{13} \times \frac{1}{17} = \frac{1}{221}\).
Q5: Which of the following represents two independent events? (a) Drawing two cards from a deck without replacement (b) Selecting a student from class A, then selecting a student from class B (c) Choosing a marble from a bag, then choosing another marble from the same bag without replacement (d) The probability of rain today and the probability that you bring an umbrella
Solution:
Ans: (b) Explanation:Independent events occur when the outcome of one event does not affect the outcome of the other. Selecting from two different classes (Class A and Class B) are independent because the selection from one class does not affect the selection from the other. Options (a) and (c) involve sampling without replacement, making them dependent. Option (d) involves events that likely influence each other.
Q6: If events A and B are independent, and \(P(A) = 0.4\) and \(P(B) = 0.3\), what is \(P(A \text{ and } B)\)? (a) 0.7 (b) 0.12 (c) 0.1 (d) 0.5
Solution:
Ans: (b) Explanation: For independent events, the probability of both events occurring is the product of their individual probabilities: \(P(A \text{ and } B) = P(A) \times P(B) = 0.4 \times 0.3 = 0.12\).
Q7: A box contains 4 green balls and 6 yellow balls. A ball is selected, its color is noted, and it is replaced. Then a second ball is selected. What is the probability of selecting a green ball both times? (a) \(\frac{2}{5}\) (b) \(\frac{4}{25}\) (c) \(\frac{12}{25}\) (d) \(\frac{1}{5}\)
Solution:
Ans: (b) Explanation: Since the ball is replaced, the events are independent. The probability of selecting a green ball on the first draw is \(\frac{4}{10} = \frac{2}{5}\). The probability of selecting a green ball on the second draw is also \(\frac{2}{5}\). Therefore: \(P(\text{green and green}) = \frac{2}{5} \times \frac{2}{5} = \frac{4}{25}\).
Q8: To determine if events A and B are independent, which condition must be satisfied? (a) \(P(A \text{ and } B) = P(A) + P(B)\) (b) \(P(A \text{ and } B) = P(A) \times P(B)\) (c) \(P(A \text{ or } B) = P(A) \times P(B)\) (d) \(P(A) = P(B)\)
Solution:
Ans: (b) Explanation: Two events are independent if and only if the probability of both events occurring equals the product of their individual probabilities: \(P(A \text{ and } B) = P(A) \times P(B)\). This is the multiplication rule for independent events.
Section B: Fill in the Blanks
Q9: Two events are called __________ if the occurrence of one event does not affect the probability of the occurrence of the other event.
Solution:
Ans: independent Explanation:Independent events are characterized by the fact that the outcome of one event has no influence on the probability of the other event occurring.
Q10: When sampling without replacement, the events are typically __________ events.
Solution:
Ans: dependent Explanation: In sampling without replacement, each selection changes the composition of the population, which affects the probabilities of subsequent selections, making the events dependent.
Q11: For two independent events A and B, the formula for the probability of both events occurring is \(P(A \text{ and } B) = \) __________.
Solution:
Ans: \(P(A) \times P(B)\) Explanation: The multiplication rule for independent events states that the probability of both events occurring is the product of their individual probabilities.
Q12: If \(P(A) = 0.6\), \(P(B) = 0.5\), and events A and B are independent, then \(P(A \text{ and } B) = \) __________.
Solution:
Ans: 0.3 Explanation: Using the multiplication rule for independent events: \(P(A \text{ and } B) = P(A) \times P(B) = 0.6 \times 0.5 = 0.3\).
Q13: When a coin is flipped and a die is rolled simultaneously, these two events are considered __________ because one does not affect the other.
Solution:
Ans: independent Explanation: Flipping a coin and rolling a die are independent events because the outcome of the coin flip has no impact on the outcome of the die roll, and vice versa.
Q14: If drawing one card from a deck affects the probability of drawing a second card, the two draws are __________ events.
Solution:
Ans: dependent Explanation: When the first card is not replaced, it changes the total number of cards and possibly the number of desired outcomes, making the second draw probability different. This makes the events dependent.
Section C: Word Problems
Q15: Sarah flips a fair coin and rolls a standard six-sided die. What is the probability that she gets tails on the coin and a number greater than 4 on the die?
Solution:
Ans: The probability of getting tails on the coin is \(\frac{1}{2}\). The numbers greater than 4 on a die are 5 and 6, so the probability is \(\frac{2}{6} = \frac{1}{3}\). Since these are independent events, we multiply the probabilities: \(P(\text{tails and number} > 4) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\) Final Answer: \(\frac{1}{6}\)
Q16: A jar contains 8 chocolate cookies and 12 vanilla cookies. If two cookies are selected one after the other without replacement, what is the probability that both cookies are chocolate?
Solution:
Ans: Total cookies initially = 8 + 12 = 20 Probability of first cookie being chocolate = \(\frac{8}{20} = \frac{2}{5}\) After removing one chocolate cookie, there are 7 chocolate cookies left out of 19 total cookies. Probability of second cookie being chocolate = \(\frac{7}{19}\) Since these are dependent events, we multiply: \(P(\text{both chocolate}) = \frac{2}{5} \times \frac{7}{19} = \frac{14}{95}\) Final Answer: \(\frac{14}{95}\)
Q17: Marcus has two bags. Bag A contains 3 red balls and 2 blue balls. Bag B contains 4 red balls and 1 blue ball. He randomly selects one ball from each bag. What is the probability that both balls selected are red?
Solution:
Ans: Bag A: Probability of selecting a red ball = \(\frac{3}{5}\) Bag B: Probability of selecting a red ball = \(\frac{4}{5}\) Since the selections from different bags are independent events: \(P(\text{both red}) = \frac{3}{5} \times \frac{4}{5} = \frac{12}{25}\) Final Answer: \(\frac{12}{25}\)
Q18: A spinner is divided into 4 equal sections numbered 1, 2, 3, and 4. The spinner is spun twice. What is the probability of landing on an even number on the first spin and an odd number on the second spin?
Solution:
Ans: Even numbers on the spinner: 2 and 4 (2 out of 4 sections) Probability of even on first spin = \(\frac{2}{4} = \frac{1}{2}\) Odd numbers on the spinner: 1 and 3 (2 out of 4 sections) Probability of odd on second spin = \(\frac{2}{4} = \frac{1}{2}\) The two spins are independent events: \(P(\text{even then odd}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\) Final Answer: \(\frac{1}{4}\)
Q19: A standard deck of 52 cards is shuffled. A card is drawn, noted, and replaced. Then the deck is shuffled again and another card is drawn. What is the probability that both cards are kings?
Solution:
Ans: There are 4 kings in a deck of 52 cards. Probability of drawing a king on the first draw = \(\frac{4}{52} = \frac{1}{13}\) Since the card is replaced and the deck is shuffled, the events are independent. Probability of drawing a king on the second draw = \(\frac{4}{52} = \frac{1}{13}\) \(P(\text{both kings}) = \frac{1}{13} \times \frac{1}{13} = \frac{1}{169}\) Final Answer: \(\frac{1}{169}\)
Q20: In a class, the probability that a randomly selected student plays basketball is 0.35, and the probability that a randomly selected student plays soccer is 0.40. If playing basketball and playing soccer are independent events, what is the probability that a randomly selected student plays both basketball and soccer?
Solution:
Ans: Let A = event that student plays basketball, with \(P(A) = 0.35\) Let B = event that student plays soccer, with \(P(B) = 0.40\) Since the events are independent: \(P(\text{both basketball and soccer}) = P(A) \times P(B) = 0.35 \times 0.40 = 0.14\) Final Answer: 0.14
The document Worksheet (with Solutions): Dependent and Independent Events is a part of the Grade 9 Course Statistics & Probability.
Semester Notes, practice quizzes, MCQs, Worksheet (with Solutions): Dependent and Independent Events, shortcuts and tricks, Worksheet (with Solutions): Dependent and Independent Events, pdf , Objective type Questions, Summary, ppt, Worksheet (with Solutions): Dependent and Independent Events, Viva Questions, Extra Questions, study material, past year papers, Sample Paper, Previous Year Questions with Solutions, mock tests for examination, Important questions, video lectures, Free, Exam;
