College Statistics Quiz: Geometric Distribution
20 questions · exam conditions
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Geometric DistributionQuestion 1 of 20

Let X be the number of flips of a biased coin needed to get the first head. The probability of heads is p. It is known that the probability of needing more than one flip is 0.64. What is the probability of needing exactly 3 flips?

0.0830.083
0.0940.094
0.1280.128
0.1470.147
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College Statistics Quiz

College Statistics Quiz: Geometric Distribution

Practice Geometric Distribution in College Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Geometric Distribution, giving you a quick way to practice the rules, question types, and explanations that matter most for College Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Let X be the number of flips of a biased coin needed to get the first head. The probability of heads is p. It is known that the probability of needing more than one flip is 0.64. What is the probability of needing exactly 3 flips?

  1. 0.0830.083
  2. 0.0940.094
  3. 0.1280.128
  4. 0.1470.147 (correct answer)
Explanation: The event 'needing more than one flip' (X>1)(X > 1) is the same as the event that the first flip is a failure (a tail). The probability of a single failure is 1−p1-p. Therefore, we are given P(X>1)=1−p=0.64P(X > 1) = 1-p = 0.64. From this, we can solve for the success probability p: p=1−0.64=0.36p = 1 - 0.64 = 0.36. The question asks for the probability of needing exactly 3 flips, which is P(X=3)P(X=3). The formula is P(X=k)=(1−p)k−1pP(X=k) = (1-p)^{k-1}p. P(X=3)=(1−0.36)3−1(0.36)=(0.64)2(0.36)=(0.4096)(0.36)≈0.147456P(X=3) = (1-0.36)^{3-1}(0.36) = (0.64)^2(0.36) = (0.4096)(0.36) \approx 0.147456.

Question 2

A gamer has a 1/3 chance of winning a prize on each independent attempt. What is the probability that the gamer's first win occurs on an odd-numbered attempt (1st, 3rd, 5th, etc.)?

  1. 2/52/5
  2. 1/31/3
  3. 13/2713/27
  4. 3/53/5 (correct answer)
Explanation: Let p=1/3p = 1/3 be the probability of winning and q=1−p=2/3q = 1-p = 2/3 be the probability of losing. The probability of the first win occurring on an odd-numbered attempt is the sum of the probabilities of winning on the 1st, 3rd, 5th, etc., attempt. P(win on 1st)=pP(\text{win on 1st}) = p P(win on 3rd)=q2pP(\text{win on 3rd}) = q^2 p P(win on 5th)=q4pP(\text{win on 5th}) = q^4 p The total probability is an infinite geometric series: p+q2p+q4p+...p + q^2 p + q^4 p + .... The first term is a=pa = p and the common ratio is r=q2r = q^2. The sum is S=a/(1−r)=p/(1−q2)S = a / (1 - r) = p / (1 - q^2). Substituting the values: S=(1/3)/(1−(2/3)2)=(1/3)/(1−4/9)=(1/3)/(5/9)=(1/3)∗(9/5)=3/5S = (1/3) / (1 - (2/3)^2) = (1/3) / (1 - 4/9) = (1/3) / (5/9) = (1/3) * (9/5) = 3/5.

Question 3

A machine has a component that fails with probability 0.1 each day. The machine starts with a working component, and the company has one identical spare. When the initial component fails, it is immediately replaced by the spare. What is the probability that the machine is non-operational (both components have failed) for the first time on the 10th day?

  1. (0.9)8(0.1)2(0.9)^8 (0.1)^2
  2. (91)(0.9)8(0.1)2\binom{9}{1} (0.9)^8 (0.1)^2 (correct answer)
  3. (0.9)9(0.1)(0.9)^9 (0.1)
  4. 1−(0.9)10−10(0.1)(0.9)91 - (0.9)^{10} - 10(0.1)(0.9)^9
Explanation: Let Y be the day the machine becomes non-operational. This is the day the second component fails. Y is the sum of the lifetimes of the two components, Y=X1+X2Y = X_1 + X_2, where X1,X2∼Geom(0.1)X_1, X_2 \sim Geom(0.1). The random variable Y follows a negative binomial distribution with r=2 and p=0.1. We want to find P(Y=10)P(Y=10). The formula for the negative binomial PMF is P(Y=k)=(k−1r−1)(1−p)k−rprP(Y=k) = \binom{k-1}{r-1} (1-p)^{k-r} p^r. Here, k=10, r=2, and p=0.1. P(Y=10)=(10−12−1)(1−0.1)10−2(0.1)2=(91)(0.9)8(0.1)2P(Y=10) = \binom{10-1}{2-1} (1-0.1)^{10-2} (0.1)^2 = \binom{9}{1} (0.9)^8 (0.1)^2. This represents the probability of having exactly one failure (the first component failing) in the first 9 days, followed by the second failure on the 10th day.

Question 4

Two archers, Alice and Bob, shoot at a target in alternating rounds, with Alice shooting first. Alice hits the bullseye with probability 0.4, and Bob hits with probability 0.3. Shots are independent. What is the probability that Alice hits the first bullseye of the game?

  1. 0.4/(0.4+0.3)0.4 / (0.4 + 0.3)
  2. 0.4/(1−(0.6)(0.7))0.4 / (1 - (0.6)(0.7)) (correct answer)
  3. 0.4+(0.6)(0.7)(0.4)0.4 + (0.6)(0.7)(0.4)
  4. 1−(0.6)(0.7)1 - (0.6)(0.7)
Explanation: Alice can win on her first shot, her second shot, her third shot, and so on. Let P(A)P(A) be the probability Alice wins.
  • Alice wins on her 1st shot (Trial 1): Alice hits. Probability = 0.4.
  • Alice wins on her 2nd shot (Trial 3): Alice misses, Bob misses, Alice hits. Probability = (0.6)(0.7)(0.4)(0.6)(0.7)(0.4).
  • Alice wins on her 3rd shot (Trial 5): (A miss, B miss), (A miss, B miss), A hits. Probability = ((0.6)(0.7))2(0.4)((0.6)(0.7))^2(0.4). This forms an infinite geometric series with first term a=0.4a = 0.4 and common ratio r=(0.6)(0.7)=0.42r = (0.6)(0.7) = 0.42. The sum of this series is given by the formula S=a/(1−r)S = a / (1 - r). P(A)=0.4/(1−0.42)=0.4/0.58P(A) = 0.4 / (1 - 0.42) = 0.4 / 0.58, which matches the expression 0.4/(1−(0.6)(0.7))0.4 / (1 - (0.6)(0.7)).

Question 5

A basketball player attempts 3-point shots until one is successful. The number of attempts required, X, follows a geometric distribution. If the variance of X is 6, what is the probability that the player makes their first successful shot on or before the third attempt?

  1. 8/278/27
  2. 4/274/27
  3. 91/21691/216
  4. 19/2719/27 (correct answer)
Explanation: The variance of a geometric distribution is Var(X)=(1−p)/p2Var(X) = (1-p)/p^2. We are given Var(X)=6Var(X) = 6. So, (1−p)/p2=6(1-p)/p^2 = 6, which gives the quadratic equation 6p2+p−1=06p^2 + p - 1 = 0. Factoring gives (3p−1)(2p+1)=0(3p-1)(2p+1)=0. Since probability p must be positive, we have p=1/3p = 1/3. The probability of failure is 1−p=2/31-p = 2/3. The question asks for the probability that the first success occurs on or before the third attempt, which is P(X≤3)P(X \le 3). The formula for the cumulative distribution function is P(X≤k)=1−(1−p)kP(X \le k) = 1 - (1-p)^k. Therefore, P(X≤3)=1−(2/3)3=1−8/27=19/27P(X \le 3) = 1 - (2/3)^3 = 1 - 8/27 = 19/27.

Question 6

A quality inspector checks products from an assembly line, with each product having a 0.2 probability of being non-conforming, independently of other products. Given that the first 10 products checked were all conforming, what is the probability that the first non-conforming product is found within the next 5 checks (i.e., on check 11, 12, 13, 14, or 15)?

  1. (0.8)5(0.8)^5
  2. (0.8)10−(0.8)15(0.8)^{10} - (0.8)^{15}
  3. 1−(0.8)51 - (0.8)^5 (correct answer)
  4. (0.8)4(0.2)(0.8)^4(0.2)
Explanation: Let X be the number of checks to find the first non-conforming product. The geometric distribution has the memoryless property, which states that P(X>a+b∣X>a)=P(X>b)P(X > a+b | X > a) = P(X > b). We are given that the first 10 products were conforming, meaning X>10X > 10. We want to find the probability that the first non-conforming product is found between checks 11 and 15, inclusive. This can be written as P(10<X≤15∣X>10)P(10 < X \le 15 | X > 10). By the memoryless property, the process effectively resets after 10 failures. We are looking for the probability that the first success occurs within the next 5 trials. This is equivalent to calculating P(X≤5)P(X \le 5) for a new geometric process. The probability of success (a non-conforming product) is p=0.2p = 0.2, and failure is 1−p=0.81-p = 0.8. P(X≤5)=1−(1−p)5=1−(0.8)5≈0.672P(X \le 5) = 1 - (1-p)^5 = 1 - (0.8)^5 \approx 0.672.

Question 7

The number of times a person takes a driving test until they pass follows a geometric distribution. The standard deviation of the number of attempts is 12\sqrt{12}. What is the probability that a person needs exactly 4 attempts to pass the test?

  1. 175/256175/256
  2. 81/102481/1024
  3. 27/25627/256 (correct answer)
  4. 1331/207361331/20736
Explanation: Let X be the number of attempts. The standard deviation is SD(X)=12SD(X) = \sqrt{12}, so the variance is Var(X)=12Var(X) = 12. The formula for the variance of a geometric distribution is Var(X)=(1−p)/p2Var(X) = (1-p)/p^2. Setting this equal to 12 gives 1−p=12p21-p = 12p^2, or 12p2+p−1=012p^2 + p - 1 = 0. Factoring yields (4p+1)(3p−1)=0(4p+1)(3p-1)=0. Since p must be positive, p=1/4=0.25p=1/4=0.25, and 1−p=3/4=0.751-p=3/4=0.75. The question asks for the probability that exactly 4 attempts are needed, which is P(X=4)P(X=4). The formula for the probability mass function is P(X=k)=(1−p)k−1pP(X=k) = (1-p)^{k-1}p. P(X=4)=(0.75)4−1(0.25)=(0.75)3(0.25)=(27/64)(1/4)=27/256P(X=4) = (0.75)^{4-1}(0.25) = (0.75)^3(0.25) = (27/64)(1/4) = 27/256.

Question 8

A biologist is searching for a rare species of frog. On any given day, the probability of finding one is 0.1, and each day's search is independent. Let Y be the number of days the biologist searches before the day they find the first frog. What is the probability that the biologist searches for at least 3 days before finding the frog?

  1. 0.07290.0729
  2. 0.8100.810
  3. 0.7290.729 (correct answer)
  4. 0.2710.271
Explanation: The variable Y is defined as the number of failures before the first success. The event 'searches for at least 3 days before finding the frog' corresponds to Y≥3Y \ge 3. This means the number of failures is 3, 4, 5, ... . This is equivalent to the first 3 days all being failures. Let p=0.1p=0.1 be the probability of success (finding a frog) and 1−p=0.91-p=0.9 be the probability of failure. The probability of the first 3 days being failures is (0.9)×(0.9)×(0.9)=(0.9)3=0.729(0.9) \times (0.9) \times (0.9) = (0.9)^3 = 0.729. Alternatively, if X is the standard geometric variable (number of trials until first success), then Y=X−1Y=X-1. The event Y≥3Y \ge 3 is equivalent to X−1≥3X-1 \ge 3, which means X≥4X \ge 4 or X>3X > 3. The probability P(X>3)=(1−p)3=(0.9)3=0.729P(X > 3) = (1-p)^3 = (0.9)^3 = 0.729.

Question 9

A machine produces items in a sequence of independent trials. The average number of items produced until the first defective one is found is 25. What is the probability that more than 30 items are produced before the first defective item is found?

  1. (0.96)29(0.04)(0.96)^{29}(0.04)
  2. 1−(0.96)301 - (0.96)^{30}
  3. (0.96)30(0.96)^{30} (correct answer)
  4. (0.04)30(0.04)^{30}
Explanation: Let X be the number of items produced until the first defective one. X follows a geometric distribution. The expected value is given as E(X)=25E(X) = 25. For a geometric distribution, E(X)=1/pE(X) = 1/p, where p is the probability of a defective item (success). Therefore, p=1/25=0.04p = 1/25 = 0.04. The probability of a non-defective item is 1−p=0.961-p = 0.96. The question asks for the probability that more than 30 items are produced, which is P(X>30)P(X > 30). This means the first 30 items must be non-defective. The formula for this is P(X>k)=(1−p)kP(X > k) = (1-p)^k. So, P(X>30)=(1−0.04)30=(0.96)30P(X > 30) = (1 - 0.04)^{30} = (0.96)^{30}.

Question 10

A student takes a 100-question multiple-choice test and guesses on every question. The probability of guessing correctly is 0.25 for each question. A researcher wants to model the number of questions the student answers until their first incorrect answer using a geometric distribution. Which of the following is the most significant reason this model may not be appropriate?

  1. The probability of an incorrect answer is not constant for each question.
  2. The trials (answering questions) are not independent of each other.
  3. There are more than two outcomes for each trial.
  4. The total number of trials is finite, not potentially infinite. (correct answer)
Explanation: The geometric distribution models the number of trials required to achieve the first success in a sequence of Bernoulli trials. A key theoretical assumption is that the trials can continue indefinitely. In this scenario, the experiment stops after 100 questions. If the student answers all 100 questions correctly, the event of a 'first incorrect answer' never occurs. This fixed upper limit on the number of trials violates the assumption of a potentially infinite sequence of trials, making the geometric distribution technically inappropriate, especially for probabilities of events near the boundary of 100.

Question 11

In a video game, the number of treasure chests a player opens to find the first rare gem follows a geometric distribution. The probability of finding the first gem within the first 4 chests is 0.5904. What is the probability of finding a rare gem in any single chest?

  1. 0.800.80
  2. 0.120.12
  3. 0.200.20 (correct answer)
  4. 0.230.23
Explanation: Let X be the number of chests opened to find the first gem, and let p be the probability of finding a gem in a single chest. We are given P(X≤4)=0.5904P(X \le 4) = 0.5904. The cumulative distribution function for a geometric distribution is P(X≤k)=1−(1−p)kP(X \le k) = 1 - (1-p)^k. So, P(X≤4)=1−(1−p)4=0.5904P(X \le 4) = 1 - (1-p)^4 = 0.5904. Rearranging the equation: (1−p)4=1−0.5904=0.4096(1-p)^4 = 1 - 0.5904 = 0.4096. To find 1−p1-p, we take the fourth root of 0.4096: 1−p=0.40964=0.81-p = \sqrt[4]{0.4096} = 0.8. Finally, we solve for p: p=1−0.8=0.2p = 1 - 0.8 = 0.2.

Question 12

A typist makes a typo on any given word with a probability of 0.05, independently. Let X be the number of words typed until the first typo. Given that the first typo occurred on or before the 10th word, what is the conditional probability that it occurred on exactly the 5th word?

  1. (0.95)4(0.05)(0.95)^4(0.05)
  2. 0.050.05
  3. (0.95)4(0.05)1−(0.95)10\frac{(0.95)^4(0.05)}{1 - (0.95)^{10}} (correct answer)
  4. 0.051−(0.95)10\frac{0.05}{1 - (0.95)^{10}}
Explanation: We need to find the conditional probability P(X=5∣X≤10)P(X=5 | X \le 10). The formula for conditional probability is P(A∣B)=P(A∩B)/P(B)P(A|B) = P(A \cap B) / P(B). Here, A is the event X=5X=5 and B is the event X≤10X \le 10. The intersection of these two events, A∩BA \cap B, is just X=5X=5. Let p=0.05p=0.05 and 1−p=0.951-p=0.95. First, calculate the numerator: P(X=5)=(1−p)5−1p=(0.95)4(0.05)P(X=5) = (1-p)^{5-1}p = (0.95)^4(0.05). Next, calculate the denominator: P(X≤10)=1−(1−p)10=1−(0.95)10P(X \le 10) = 1 - (1-p)^{10} = 1 - (0.95)^{10}. Therefore, the conditional probability is P(X=5)P(X≤10)=(0.95)4(0.05)1−(0.95)10\frac{P(X=5)}{P(X \le 10)} = \frac{(0.95)^4(0.05)}{1 - (0.95)^{10}}.

Question 13

A scientist conducts a series of independent experiments, each with a 40% chance of success. Let X be the number of experiments conducted up to and including the first success. What is the expected number of failures the scientist will have before achieving the first success?

  1. 2.52.5
  2. 1.51.5 (correct answer)
  3. 0.60.6
  4. 1.671.67
Explanation: The probability of success is p=0.4p = 0.4. The random variable X, the number of trials until the first success, has an expected value of E(X)=1/p=1/0.4=2.5E(X) = 1/p = 1/0.4 = 2.5. This represents the total number of trials (failures + 1 success). The question asks for the expected number of failures before the first success. If Y is the number of failures, then Y=X−1Y = X - 1. Using the linearity of expectation, E(Y)=E(X−1)=E(X)−1=2.5−1=1.5E(Y) = E(X - 1) = E(X) - 1 = 2.5 - 1 = 1.5. Alternatively, the expected number of failures before the first success has a direct formula of (1−p)/p=0.6/0.4=1.5(1-p)/p = 0.6/0.4 = 1.5.

Question 14

Let X1X_1 be the number of trials for the first success in a series of Bernoulli trials with success probability p. Let X2X_2 be the number of additional trials for the second success. Assume X1X_1 and X2X_2 are independent. Let Y=X1+X2Y = X_1 + X_2. Which distribution does the random variable Y follow?

  1. A geometric distribution with probability p2p^2
  2. A binomial distribution with a number of trials equal to E[X1+X2]E[X_1+X_2]
  3. A geometric distribution with probability 2p2p
  4. A negative binomial distribution for the number of trials to get 2 successes (correct answer)
Explanation: X1X_1 is the waiting time for the 1st success, and X2X_2 is the waiting time from the 1st success to the 2nd success. Their sum, Y, represents the total number of trials required to achieve exactly 2 successes. By definition, this is the negative binomial distribution with parameters r=2 (number of successes) and p (probability of success).

Question 15

A customer service center finds that 15% of callers need to be transferred to a supervisor. A new employee's calls are monitored. What is the probability that the employee's first transfer to a supervisor occurs after their 5th call but on or before their 8th call?

  1. (0.85)5−(0.85)8(0.85)^5 - (0.85)^8 (correct answer)
  2. (0.85)6−(0.85)8(0.85)^6 - (0.85)^8
  3. 1−(0.85)31 - (0.85)^3
  4. (0.85)5(0.15)+(0.85)6(0.15)+(0.85)7(0.15)(0.85)^5(0.15) + (0.85)^6(0.15) + (0.85)^7(0.15)
Explanation: Let X be the number of calls until the first transfer. We have a geometric distribution with p=0.15p=0.15. The question asks for the probability that the first transfer occurs after the 5th call and on or before the 8th call, which is P(5<X≤8)P(5 < X \le 8). This can be calculated as P(X≤8)−P(X≤5)P(X \le 8) - P(X \le 5). However, an easier method is to use the formula P(X>k)=(1−p)kP(X > k) = (1-p)^k. P(5<X≤8)=P(X>5)−P(X>8)P(5 < X \le 8) = P(X > 5) - P(X > 8). This represents the probability that the first 5 calls are not transfers, minus the probability that the first 8 calls are not transfers. The difference is the probability that the first transfer happens on call 6, 7, or 8. Let 1−p=0.851-p = 0.85. The probability is (0.85)5−(0.85)8(0.85)^5 - (0.85)^8.

Question 16

The probability that a particle passes through a filter is 0.8. Particles are fired independently at the filter. What is the minimum number of particles that must be fired so that the probability of at least one particle failing to pass through is greater than 0.99?

  1. 20
  2. 21 (correct answer)
  3. 22
  4. 23
Explanation: The probability that a particle fails to pass through is 0.2. We want the minimum number of particles k such that the probability of at least one failure in k trials is greater than 0.99. The probability of at least one failure is 1 minus the probability that all particles pass through: P(at least one failure)=1−(0.8)kP(\text{at least one failure}) = 1 - (0.8)^k. We need 1−(0.8)k>0.991 - (0.8)^k > 0.99, which gives us (0.8)k<0.01(0.8)^k < 0.01. Taking natural logarithm: kln⁡(0.8)<ln⁡(0.01)k \ln(0.8) < \ln(0.01). Since ln⁡(0.8)<0\ln(0.8) < 0, we reverse the inequality: k>ln⁡(0.01)ln⁡(0.8)=−4.605−0.223≈20.63k > \frac{\ln(0.01)}{\ln(0.8)} = \frac{-4.605}{-0.223} \approx 20.63. The smallest integer greater than 20.63 is 21.

Question 17

Let X be a geometric random variable with mean μ=5\mu = 5. What is the probability that the first success occurs on a trial number that is greater than the mean?

  1. (0.8)4(0.2)(0.8)^4(0.2)
  2. 1−(0.8)51-(0.8)^5
  3. (0.8)4(0.8)^4
  4. (0.8)5(0.8)^5 (correct answer)
Explanation: First, find the success probability p. Given the mean μ=E(X)=1/p=5\mu = E(X) = 1/p = 5, we find p=1/5=0.2p = 1/5 = 0.2. The probability of failure is 1−p=0.81-p = 0.8. The question asks for the probability that the first success occurs on a trial number greater than the mean, which is P(X>5)P(X > 5). This means the first 5 trials must all be failures. The formula for this is P(X>k)=(1−p)kP(X > k) = (1-p)^k. Substituting k=5k=5 and p=0.2p=0.2, we get P(X>5)=(1−0.2)5=(0.8)5P(X > 5) = (1 - 0.2)^5 = (0.8)^5.

Question 18

A student takes a 100-question multiple-choice test and guesses on every question. The probability of guessing correctly is 0.25 for each question. A researcher wants to model the number of questions the student answers until their first incorrect answer using a geometric distribution. Which of the following is the most significant reason this model may not be appropriate?

  1. The probability of an incorrect answer is not constant for each question.
  2. The trials (answering questions) are not independent of each other.
  3. There are more than two outcomes for each trial.
  4. The total number of trials is finite, not potentially infinite. (correct answer)
Explanation: The geometric distribution models the number of trials required to achieve the first success in a sequence of Bernoulli trials. A key theoretical assumption is that the trials can continue indefinitely. In this scenario, the experiment stops after 100 questions. If the student answers all 100 questions correctly, the event of a 'first incorrect answer' never occurs. This fixed upper limit on the number of trials violates the assumption of a potentially infinite sequence of trials, making the geometric distribution technically inappropriate, especially for probabilities of events near the boundary of 100.

Question 19

In a video game, the number of treasure chests a player opens to find the first rare gem follows a geometric distribution. The probability of finding the first gem within the first 4 chests is 0.5904. What is the probability of finding a rare gem in any single chest?

  1. 0.800.80
  2. 0.120.12
  3. 0.200.20 (correct answer)
  4. 0.230.23
Explanation: Let X be the number of chests opened to find the first gem, and let p be the probability of finding a gem in a single chest. We are given P(X≤4)=0.5904P(X \le 4) = 0.5904. The cumulative distribution function for a geometric distribution is P(X≤k)=1−(1−p)kP(X \le k) = 1 - (1-p)^k. So, P(X≤4)=1−(1−p)4=0.5904P(X \le 4) = 1 - (1-p)^4 = 0.5904. Rearranging the equation: (1−p)4=1−0.5904=0.4096(1-p)^4 = 1 - 0.5904 = 0.4096. To find 1−p1-p, we take the fourth root of 0.4096: 1−p=0.40964=0.81-p = \sqrt[4]{0.4096} = 0.8. Finally, we solve for p: p=1−0.8=0.2p = 1 - 0.8 = 0.2.

Question 20

Let X be a random variable following a geometric distribution with success probability p. Which of the following statements correctly describes the shape of the probability mass function of X?

  1. The distribution is symmetric if p=0.5p=0.5 and skewed otherwise.
  2. The shape of the distribution approaches a Normal distribution as the expected value increases.
  3. The distribution is always skewed to the left.
  4. The distribution is always skewed to the right, with the mode at 1. (correct answer)
Explanation: For a geometric distribution, the probability of the first success occurring on the k-th trial is P(X=k)=(1−p)k−1pP(X=k) = (1-p)^{k-1}p. The most likely outcome is always the first trial (k=1), with probability p. As k increases, the term (1−p)k−1(1-p)^{k-1} decreases, so P(X=k)P(X=k) is a strictly decreasing function of k. This creates a long tail to the right, meaning the distribution is always right-skewed, regardless of the value of p. The mode (the most frequent value) is always at X=1.