Q1: A geometric random variable counts the number of trials needed to get the first success. If the probability of success on each trial is \(p = 0.3\), what is the probability that the first success occurs on the 4th trial? (a) 0.1029 (b) 0.3 (c) 0.0441 (d) 0.7
Solution:
Ans: (a) Explanation: For a geometric distribution, the probability that the first success occurs on the \(n\)th trial is \(P(X = n) = (1-p)^{n-1} \cdot p\). Here, \(P(X = 4) = (0.7)^3 \cdot 0.3 = 0.343 \cdot 0.3 = 0.1029\). Option (b) is just \(p\), option (c) is \((0.7)^4\), and option (d) is \(1-p\).
Q2: Which of the following is NOT a requirement for a geometric random variable? (a) Each trial has two possible outcomes: success or failure (b) The probability of success is the same for each trial (c) The trials are independent (d) The number of trials is fixed in advance
Solution:
Ans: (d) Explanation: A geometric random variable counts trials until the first success, so the number of trials is not fixed in advance. Options (a), (b), and (c) are all required conditions for a geometric setting. A fixed number of trials describes a binomial distribution, not geometric.
Q3: The mean (expected value) of a geometric random variable with probability of success \(p\) is given by which formula? (a) \(\frac{1}{p}\) (b) \(np\) (c) \(\frac{1-p}{p}\) (d) \(p(1-p)\)
Solution:
Ans: (a) Explanation: The expected value of a geometric random variable is \(\mu = \frac{1}{p}\), representing the average number of trials needed to get the first success. Option (b) is the mean of a binomial distribution, option (c) relates to variance components, and option (d) is not a standard formula for geometric distributions.
Q4: A basketball player makes free throws with probability 0.8. What is the probability that she makes her first free throw on or before the 3rd attempt? (a) 0.992 (b) 0.8 (c) 0.896 (d) 0.512
Q5: The standard deviation of a geometric random variable with \(p = 0.25\) is closest to: (a) 3.46 (b) 4 (c) 12 (d) 0.75
Solution:
Ans: (a) Explanation: The standard deviation of a geometric random variable is \(\sigma = \frac{\sqrt{1-p}}{p}\). Here, \(\sigma = \frac{\sqrt{0.75}}{0.25} = \frac{0.866}{0.25} = 3.464\). Option (b) is the mean \(\frac{1}{p}\), option (c) is the variance, and option (d) is \(1-p\).
Q6: If \(X\) follows a geometric distribution with \(p = 0.4\), what is \(P(X > 2)\)? (a) 0.36 (b) 0.64 (c) 0.6 (d) 0.24
Solution:
Ans: (a) Explanation: For a geometric random variable, \(P(X > n) = (1-p)^n\). Thus, \(P(X > 2) = (0.6)^2 = 0.36\). This represents the probability that the first two trials are both failures. Option (b) is \(1-p\), option (c) is just \(1-p\), and option (d) is \(0.6 \cdot 0.4\).
Q7: A quality control inspector tests items until finding a defective one. If 5% of items are defective, what is the expected number of items that must be tested? (a) 20 (b) 5 (c) 0.05 (d) 19
Solution:
Ans: (a) Explanation: The expected value is \(\mu = \frac{1}{p} = \frac{1}{0.05} = 20\). On average, the inspector needs to test 20 items to find the first defective one. Option (b) is the reciprocal, option (c) is \(p\) itself, and option (d) is \(\frac{1-p}{p}\).
Q8: Which probability distribution is memoryless, meaning that \(P(X > s + t | X > s) = P(X > t)\)? (a) Geometric distribution (b) Binomial distribution (c) Normal distribution (d) Uniform distribution
Solution:
Ans: (a) Explanation: The geometric distribution has the memoryless property, which means that the probability of success in future trials does not depend on how many failures have already occurred. This property is unique to geometric (and exponential) distributions among those listed.
## Section B: Fill in the Blanks Q9: The probability mass function for a geometric random variable is \(P(X = n) = (1-p)^{n-1} \cdot p\), where \(n\) represents the trial on which the __________ success occurs.
Solution:
Ans: first Explanation: A geometric random variable specifically counts the number of trials until the first success is achieved. This distinguishes it from other discrete distributions.
Q10: The variance of a geometric random variable with probability of success \(p\) is given by the formula __________.
Solution:
Ans: \(\frac{1-p}{p^2}\) Explanation: The variance formula for a geometric distribution is \(\sigma^2 = \frac{1-p}{p^2}\), which measures the spread of the distribution around the mean.
Q11: For a geometric random variable, the trials must be __________, meaning the outcome of one trial does not affect another.
Solution:
Ans: independent Explanation:Independence is a crucial assumption for geometric distributions. Each trial must have the same probability of success regardless of previous outcomes.
Q12: If a geometric random variable has \(p = 0.2\), then the probability that the first success occurs after the 5th trial is __________.
Solution:
Ans: 0.32768 or \((0.8)^5\) Explanation: \(P(X > 5) = (1-p)^5 = (0.8)^5 = 0.32768\). This represents the probability that all of the first 5 trials are failures.
Q13: In a geometric setting, each trial must result in one of two outcomes: success or __________.
Solution:
Ans: failure Explanation: A geometric random variable requires a binary outcome for each trial: either success or failure. This is one of the key conditions for a geometric distribution.
Q14: If the mean of a geometric distribution is 8, then the probability of success on each trial is __________.
Solution:
Ans: 0.125 or \(\frac{1}{8}\) Explanation: Since \(\mu = \frac{1}{p} = 8\), we solve for \(p\): \(p = \frac{1}{8} = 0.125\). The mean and probability of success are reciprocals in a geometric distribution.
## Section C: Word Problems Q15: A student guesses on a multiple-choice quiz where each question has 4 options and only one correct answer. Assuming the student guesses randomly on each question, what is the probability that the student gets the first correct answer on the 5th question?
Solution:
Ans: Step 1: The probability of success (correct answer) is \(p = \frac{1}{4} = 0.25\). Step 2: The probability of failure is \(1 - p = 0.75\). Step 3: Using the geometric probability formula: \(P(X = 5) = (0.75)^4 \cdot 0.25\). Step 4: Calculate: \((0.75)^4 = 0.31640625\). Step 5: \(0.31640625 \cdot 0.25 = 0.0791015625\). Final Answer: Approximately 0.0791 or 7.91%
Q16: A factory produces light bulbs, and 2% are defective. An inspector tests bulbs randomly one at a time until finding a defective one. What is the expected number of bulbs the inspector will need to test? What is the standard deviation?
Solution:
Ans: Step 1: The probability of finding a defective bulb is \(p = 0.02\). Step 2: The expected value is \(\mu = \frac{1}{p} = \frac{1}{0.02} = 50\) bulbs. Step 3: The variance is \(\sigma^2 = \frac{1-p}{p^2} = \frac{0.98}{(0.02)^2} = \frac{0.98}{0.0004} = 2450\). Step 4: The standard deviation is \(\sigma = \sqrt{2450} \approx 49.497\) bulbs. Final Answer: Expected value = 50 bulbs; Standard deviation ≈ 49.5 bulbs
Q17: A archer hits the bullseye 60% of the time. What is the probability that the archer's first bullseye occurs on the 3rd shot or earlier?
Solution:
Ans: Step 1: The probability of success is \(p = 0.6\). Step 2: We need \(P(X \leq 3) = 1 - P(X > 3)\). Step 3: \(P(X > 3) = (1-p)^3 = (0.4)^3 = 0.064\). Step 4: \(P(X \leq 3) = 1 - 0.064 = 0.936\). Final Answer: 0.936 or 93.6%
Q18: A baseball player has a batting average of 0.300, meaning he gets a hit 30% of the time at bat. What is the probability that he gets his first hit on exactly his 4th at-bat? What is the probability it takes him more than 4 at-bats to get his first hit?
Solution:
Ans: Part 1 - First hit on 4th at-bat: Step 1: \(p = 0.3\) and \(1-p = 0.7\). Step 2: \(P(X = 4) = (0.7)^3 \cdot 0.3 = 0.343 \cdot 0.3 = 0.1029\). Part 2 - More than 4 at-bats: Step 3: \(P(X > 4) = (0.7)^4 = 0.2401\). Final Answer: \(P(X = 4) = 0.1029\) or 10.29%; \(P(X > 4) = 0.2401\) or 24.01%
Q19: A telemarketer successfully makes a sale with probability 0.08 on each call. Calls are independent. If the telemarketer's goal is to make one sale, what is the probability that this goal is achieved within the first 10 calls?
Solution:
Ans: Step 1: The probability of success is \(p = 0.08\). Step 2: We need \(P(X \leq 10) = 1 - P(X > 10)\). Step 3: \(P(X > 10) = (1-p)^{10} = (0.92)^{10}\). Step 4: Calculate: \((0.92)^{10} \approx 0.4344\). Step 5: \(P(X \leq 10) = 1 - 0.4344 = 0.5656\). Final Answer: Approximately 0.5656 or 56.56%
Q20: A game involves rolling a fair six-sided die until a 6 appears. Using the geometric distribution model, find the probability that it takes exactly 3 rolls to get the first 6, and calculate the expected number of rolls needed.
Solution:
Ans: Part 1 - Exactly 3 rolls: Step 1: The probability of rolling a 6 is \(p = \frac{1}{6}\). Step 2: The probability of not rolling a 6 is \(1 - p = \frac{5}{6}\). Step 3: \(P(X = 3) = \left(\frac{5}{6}\right)^2 \cdot \frac{1}{6} = \frac{25}{36} \cdot \frac{1}{6} = \frac{25}{216} \approx 0.1157\). Part 2 - Expected rolls: Step 4: \(\mu = \frac{1}{p} = \frac{1}{\frac{1}{6}} = 6\) rolls. Final Answer: \(P(X = 3) = \frac{25}{216} \approx 0.1157\) or 11.57%; Expected rolls = 6
The document Worksheet (with Solutions): Geometric Random Variables is a part of the Grade 9 Course Statistics & Probability.
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