Discrete Math Quiz: Geometric And Negative Binomial Distributions
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Geometric And Negative Binomial DistributionsQuestion 1 of 19

Two independent geometric random variables XX and YY both have parameter p=0.3p = 0.3. What is P(X=Y=5)P(X = Y = 5)?

(21)(0.3)(0.7)4\binom{2}{1}(0.3)(0.7)^4
(0.3)2(0.7)8(0.3)^2(0.7)^8
[(0.3)(0.7)4]2[(0.3)(0.7)^4]^2
(0.3)(0.7)4+(0.3)(0.7)4(0.3)(0.7)^4 + (0.3)(0.7)^4
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Discrete Math Quiz

Discrete Math Quiz: Geometric And Negative Binomial Distributions

Practice Geometric And Negative Binomial Distributions in Discrete Math 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 And Negative Binomial Distributions, giving you a quick way to practice the rules, question types, and explanations that matter most for Discrete Math.

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

Two independent geometric random variables XX and YY both have parameter p=0.3p = 0.3. What is P(X=Y=5)P(X = Y = 5)?

  1. (21)(0.3)(0.7)4\binom{2}{1}(0.3)(0.7)^4
  2. (0.3)2(0.7)8(0.3)^2(0.7)^8
  3. [(0.3)(0.7)4]2[(0.3)(0.7)^4]^2 (correct answer)
  4. (0.3)(0.7)4+(0.3)(0.7)4(0.3)(0.7)^4 + (0.3)(0.7)^4
Explanation: When you encounter probability questions involving independent random variables, remember that independence means you can multiply their individual probabilities to find joint probabilities. For a geometric random variable with parameter pp, the probability of the first success occurring on trial nn is P(X=n)=(1−p)n−1⋅pP(X = n) = (1-p)^{n-1} \cdot p. With p=0.3p = 0.3, we have P(X=5)=(0.7)4⋅(0.3)P(X = 5) = (0.7)^4 \cdot (0.3), since we need 4 failures followed by 1 success. Since XX and YY are independent, P(X=Y=5)=P(X=5)⋅P(Y=5)P(X = Y = 5) = P(X = 5) \cdot P(Y = 5). Each variable has the same probability: (0.3)(0.7)4(0.3)(0.7)^4. Therefore, P(X=Y=5)=[(0.3)(0.7)4]2P(X = Y = 5) = [(0.3)(0.7)^4]^2, which is answer C. Let's examine why the other options are incorrect: Option A, (21)(0.3)(0.7)4\binom{2}{1}(0.3)(0.7)^4, mistakenly applies binomial coefficients, as if we're choosing which of two trials succeeds. This isn't a binomial problem—both variables must equal 5. Option B, (0.3)2(0.7)8(0.3)^2(0.7)^8, correctly identifies that we need two successes total but incorrectly distributes the trials, treating this as if we have 10 total trials with 2 successes anywhere. Option D, (0.3)(0.7)4+(0.3)(0.7)4(0.3)(0.7)^4 + (0.3)(0.7)^4, adds probabilities instead of multiplying them, confusing "and" with "or." Remember: for independent events occurring simultaneously, multiply their individual probabilities. The word "and" in probability typically signals multiplication when events are independent.

Question 2

Two independent geometric random variables X and Y both have the same success probability p = 0.3. What is the probability that X + Y = 5?

  1. 0.73×0.32×30.7^3 \times 0.3^2 \times 3
  2. 0.73×0.32×40.7^3 \times 0.3^2 \times 4 (correct answer)
  3. 0.74×0.32×40.7^4 \times 0.3^2 \times 4
  4. 0.72×0.33×40.7^2 \times 0.3^3 \times 4
Explanation: When you encounter problems involving sums of independent geometric random variables, you need to consider all possible ways the sum can occur and apply the independence property. For geometric random variables with success probability p = 0.3, the probability of the first success on trial k is (1−p)k−1⋅p=0.7k−1⋅0.3(1-p)^{k-1} \cdot p = 0.7^{k-1} \cdot 0.3. To find P(X + Y = 5), you must identify all combinations where X and Y sum to 5: (1,4), (2,3), (3,2), and (4,1). Since X and Y are independent, you calculate each probability separately and sum them:
  • P(X=1, Y=4) = 0.3×0.73×0.3=0.73×0.320.3 \times 0.7^3 \times 0.3 = 0.7^3 \times 0.3^2
  • P(X=2, Y=3) = 0.7×0.3×0.72×0.3=0.73×0.320.7 \times 0.3 \times 0.7^2 \times 0.3 = 0.7^3 \times 0.3^2
  • P(X=3, Y=2) = 0.72×0.3×0.7×0.3=0.73×0.320.7^2 \times 0.3 \times 0.7 \times 0.3 = 0.7^3 \times 0.3^2
  • P(X=4, Y=1) = 0.73×0.3×0.3=0.73×0.320.7^3 \times 0.3 \times 0.3 = 0.7^3 \times 0.3^2
All four scenarios have the same probability, so the total is 4×0.73×0.324 \times 0.7^3 \times 0.3^2, making answer B correct. Answer A uses 3 instead of 4 scenarios, missing one valid combination. Answer C incorrectly uses 0.740.7^4, suggesting a miscalculation in the exponents. Answer D uses 0.72×0.330.7^2 \times 0.3^3, which doesn't match the geometric distribution formula for any valid combination. Remember: when finding probabilities for sums of discrete random variables, systematically list all possible combinations and count them carefully—this counting step is where most errors occur.

Question 3

A machine produces items with a 0.15 defect rate. An inspector checks items until finding 2 defective ones, then stops. If the inspection process is repeated many times, what percentage of the time will the inspector examine exactly 10 items?

  1. 10×(0.85)8×(0.15)2×100%10 \times (0.85)^8 \times (0.15)^2 \times 100\%
  2. 9×(0.85)8×(0.15)2×100%9 \times (0.85)^8 \times (0.15)^2 \times 100\% (correct answer)
  3. (102)×(0.85)8×(0.15)2×100%\binom{10}{2} \times (0.85)^8 \times (0.15)^2 \times 100\%
  4. 8×(0.85)9×(0.15)2×100%8 \times (0.85)^9 \times (0.15)^2 \times 100\%
Explanation: When you encounter a problem about finding a specific number of successes before stopping, you're dealing with a negative binomial distribution. The key insight is recognizing the constraint: the process must end on exactly the target item. Here's the scenario: the inspector stops immediately after finding the 2nd defective item. So if exactly 10 items are examined, the 10th item must be defective (the 2nd defective one), and among the first 9 items, exactly 1 must be defective. The probability breaks down as:
  • Ways to choose which of the first 9 items is defective: 99 ways
  • Probability those 9 items have exactly 1 defective: (0.15)1×(0.85)8(0.15)^1 \times (0.85)^8
  • Probability the 10th item is defective: 0.150.15
This gives us 9×(0.85)8×(0.15)29 \times (0.85)^8 \times (0.15)^2, which matches answer B. Let's examine why the other options fail: A uses coefficient 10 instead of 9, incorrectly assuming you can choose any of the 10 positions for a defective item. But the 10th position is fixed—it must be defective. C uses (102)\binom{10}{2}, treating this like a standard binomial problem where you'd choose 2 positions out of 10 for defective items. This ignores the stopping condition. D uses coefficient 8 and (0.85)9(0.85)^9, suggesting 8 ways to arrange items among 9 positions, which doesn't match the constraint structure. Study tip: In negative binomial problems, always remember that the final trial is fixed as a success, so you only arrange the earlier trials.

Question 4

A basketball player practices free throws until making 5 successful shots. Her probability of making any single shot is 0.70.7. If she has already missed 8 shots and made 4 successful shots, what is the probability that she will need exactly 3 more attempts to complete her practice session?

  1. (0.3)2⋅(0.7)(0.3)^2 \cdot (0.7) (correct answer)
  2. (0.7)2⋅(0.3)(0.7)^2 \cdot (0.3)
  3. (20)⋅(0.3)2⋅(0.7)\binom{2}{0} \cdot (0.3)^2 \cdot (0.7)
  4. (31)⋅(0.7)⋅(0.3)2\binom{3}{1} \cdot (0.7) \cdot (0.3)^2
Explanation: She needs 1 more successful shot. Given that she needs exactly 3 more attempts, she must miss the next 2 shots and make the 3rd. This probability is (0.3)2⋅(0.7)(0.3)^2 \cdot (0.7). The memoryless property means past attempts don't affect future probabilities. Choice B reverses the probabilities. Choice C includes an unnecessary binomial coefficient. Choice D incorrectly suggests multiple ways to arrange the shots.

Question 5

A software testing team runs automated tests until they find their first critical bug. Each test has a probability of 0.150.15 of finding a critical bug, independent of other tests. If the expected number of tests needed is 203\frac{20}{3}, what is the probability that they will need to run more than 8 tests to find their first critical bug?

  1. (0.85)8(0.85)^8 (correct answer)
  2. (0.85)9(0.85)^9
  3. 1−(0.85)81 - (0.85)^8
  4. 8⋅(0.15)⋅(0.85)78 \cdot (0.15) \cdot (0.85)^7
Explanation: This is a geometric distribution with p=0.15p = 0.15. We need P(X>8)P(X > 8) where XX is the number of tests. For a geometric distribution, P(X>k)=(1−p)kP(X > k) = (1-p)^k. So P(X>8)=(0.85)8P(X > 8) = (0.85)^8. Choice B gives P(X>9)P(X > 9). Choice C gives P(X≤8)P(X \leq 8). Choice D gives P(X=8)P(X = 8).

Question 6

A computer network experiences random failures. The probability of failure on any given day is 0.120.12. What is the expected number of days until the network experiences its second failure, starting from a functioning state?

  1. 20.122=12509\frac{2}{0.12^2} = \frac{1250}{9}
  2. 10.12+10.12=503\frac{1}{0.12} + \frac{1}{0.12} = \frac{50}{3}
  3. 2−10.12=253\frac{2-1}{0.12} = \frac{25}{3}
  4. 20.12=503\frac{2}{0.12} = \frac{50}{3} (correct answer)
Explanation: When you encounter problems about waiting for the "r-th occurrence" of an event, you're dealing with negative binomial distributions. The key insight is recognizing what "expected number of days until the second failure" actually means. This problem asks for the expected value of the random variable representing the day on which the second failure occurs. For a negative binomial distribution where we wait for the r-th success with probability p of success on each trial, the expected value is rp\frac{r}{p}. Here, we want the second failure (r = 2) with failure probability p = 0.12, giving us 20.12=503\frac{2}{0.12} = \frac{50}{3}. Choice A (20.122=12509\frac{2}{0.12^2} = \frac{1250}{9}) incorrectly squares the probability. This might come from confusing the formula with variance calculations or thinking about the probability of two specific failures occurring. Choice B (10.12+10.12=503\frac{1}{0.12} + \frac{1}{0.12} = \frac{50}{3}) represents a common misconception: adding the expected waiting times for the first failure and then the second failure. While this gives the same numerical answer as D, the reasoning is flawed because it treats these as independent waiting periods rather than a single process. Choice C (2−10.12=253\frac{2-1}{0.12} = \frac{25}{3}) uses r-1 instead of r, which would give the expected number of additional days needed after the first failure, not the total days from the beginning. Remember: for "time until r-th occurrence" problems, use rp\frac{r}{p} directly. Don't overthink by breaking it into separate waiting periods.

Question 7

A researcher conducts trials until observing exactly 6 successes. Each trial has success probability 0.250.25. If the researcher stops after exactly 18 trials, what was the probability of success on the 10th trial, given the final outcome?

  1. 618=13\frac{6}{18} = \frac{1}{3}
  2. 0.250.25 (correct answer)
  3. 517\frac{5}{17}
  4. 6⋅0.2518=112\frac{6 \cdot 0.25}{18} = \frac{1}{12}
Explanation: When you encounter problems about conditional probability in scenarios where trials have already been completed, the key insight is recognizing what information actually changes given the outcome. The correct answer is B (0.250.25) because each trial's probability of success remains independent and unchanged, regardless of the final outcome. Even though we know the researcher observed exactly 6 successes in 18 trials, this knowledge doesn't alter the fundamental probability that any specific trial (including the 10th) was a success. Each trial still had a 0.250.25 probability of success when it occurred. Let's examine why the other options represent common misconceptions. Choice A (13\frac{1}{3}) incorrectly assumes that knowing the total success rate (6 successes out of 18 trials) somehow becomes the probability for individual trials. This confuses the observed frequency with the underlying probability. Choice C (517\frac{5}{17}) appears to use some form of conditional reasoning about the remaining trials, but this misapplies conditional probability concepts. Choice D (112\frac{1}{12}) seems to blend the individual trial probability with the total number of trials, creating a meaningless calculation. The trap here is thinking that additional information about the final outcome changes the probability of past events. Remember: when dealing with independent trials, the probability of any specific trial having a particular outcome equals the original trial probability, regardless of what you learn about other trials afterward. Focus on whether the trials are independent rather than being swayed by aggregate outcomes.

Question 8

A game involves rolling a fair die until getting exactly 3 sixes. Let TT be the number of rolls needed. If the probability generating function of TT is (z/61−5z/6)3\left(\frac{z/6}{1-5z/6}\right)^3, what is P(T=5)P(T = 5)?

  1. (53)(16)3(56)2\binom{5}{3} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^2
  2. (42)(16)3(56)2\binom{4}{2} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^2 (correct answer)
  3. 3(16)3(56)23 \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^2
  4. (16)3(56)2\left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^2
Explanation: This problem involves the negative binomial distribution, which models the number of trials needed to achieve a fixed number of successes. When you see a probability generating function (PGF) of the form (pz1−(1−p)z)r\left(\frac{pz}{1-(1-p)z}\right)^r, you're dealing with the distribution of trials needed to get exactly rr successes. To find P(T=5)P(T = 5), you need the coefficient of z5z^5 in the given PGF. However, there's a more direct approach using the negative binomial probability formula. Since we need exactly 3 sixes in 5 rolls, the last roll must be a six (to reach exactly 3 sixes), and among the first 4 rolls, exactly 2 must be sixes. The probability is: (42)(16)3(56)2\binom{4}{2} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^2 Here's why: (42)\binom{4}{2} counts the ways to choose which 2 of the first 4 rolls are sixes, (16)3\left(\frac{1}{6}\right)^3 is the probability of getting 3 sixes total, and (56)2\left(\frac{5}{6}\right)^2 is the probability of getting 2 non-sixes. A is wrong because (53)\binom{5}{3} would allow any arrangement of 3 sixes in 5 rolls, but we need the 5th roll to be the final six. C incorrectly uses 3 instead of the proper binomial coefficient. D omits the combinatorial factor entirely, ignoring that there are multiple ways to arrange the first 2 sixes among 4 rolls. Study tip: For "trials until rrth success" problems, remember the last trial must be a success, so you're really choosing positions for the other (r−1)(r-1) successes among the previous trials.

Question 9

A coin has probability pp of showing heads. An experiment consists of flipping the coin until getting 3 heads total. If the variance of the number of flips needed is 3(1−p)p2\frac{3(1-p)}{p^2}, what is the probability that exactly 7 flips are needed when p=0.4p = 0.4?

  1. (52)⋅(0.4)3⋅(0.6)4\binom{5}{2} \cdot (0.4)^3 \cdot (0.6)^4
  2. (73)⋅(0.4)3⋅(0.6)4\binom{7}{3} \cdot (0.4)^3 \cdot (0.6)^4
  3. (63)⋅(0.4)3⋅(0.6)3\binom{6}{3} \cdot (0.4)^3 \cdot (0.6)^3
  4. (62)⋅(0.4)3⋅(0.6)4\binom{6}{2} \cdot (0.4)^3 \cdot (0.6)^4 (correct answer)
Explanation: This is a negative binomial distribution problem, where you're flipping until you get a fixed number of successes (3 heads). The key insight is understanding what "exactly 7 flips" means: the 7th flip must be heads (the 3rd head), and among the first 6 flips, exactly 2 must be heads. To find the probability of exactly 7 flips needed, you need to calculate: (ways to get 2 heads in first 6 flips) × (probability of those arrangements) × (probability 7th flip is heads). This gives us (62)⋅(0.4)2⋅(0.6)4⋅(0.4)1=(62)⋅(0.4)3⋅(0.6)4\binom{6}{2} \cdot (0.4)^2 \cdot (0.6)^4 \cdot (0.4)^1 = \binom{6}{2} \cdot (0.4)^3 \cdot (0.6)^4, which is answer D. Looking at the wrong answers: A uses (52)\binom{5}{2}, which incorrectly assumes you're choosing from only 5 flips instead of the first 6. B uses (73)\binom{7}{3}, treating this like a regular binomial problem where any 3 of 7 flips could be heads, ignoring that the final flip must be heads. C uses (0.6)3(0.6)^3, suggesting only 3 tails total, but if you need exactly 7 flips with 3 heads, you must have 4 tails. Remember: in "flips until r successes" problems, the final flip is always a success, so you're really arranging the first (r-1) successes among the first (n-1) trials. The pattern is (n−1r−1)⋅pr⋅(1−p)n−r\binom{n-1}{r-1} \cdot p^r \cdot (1-p)^{n-r}.

Question 10

A manufacturing process produces defective items with probability 0.050.05. Items are tested sequentially until finding the first defective item. Given that more than 10 items have been tested, what is the probability that more than 20 items total will be tested?

  1. (0.95)20(0.95)10\frac{(0.95)^{20}}{(0.95)^{10}}
  2. (0.95)20(0.95)^{20}
  3. (0.95)10(0.95)^{10} (correct answer)
  4. (0.95)10(0.95)20\frac{(0.95)^{10}}{(0.95)^{20}}
Explanation: This question tests conditional probability in the context of a geometric distribution, where you're looking for the first success (defective item) in a sequence of independent trials. The key insight is recognizing the memoryless property of geometric distributions. You need to find P(more than 20 tested∣more than 10 tested)P(\text{more than 20 tested} | \text{more than 10 tested}). Using conditional probability: P(X>20∣X>10)=P(X>20 and X>10)P(X>10)=P(X>20)P(X>10)P(X > 20 | X > 10) = \frac{P(X > 20 \text{ and } X > 10)}{P(X > 10)} = \frac{P(X > 20)}{P(X > 10)} For a geometric distribution with success probability p=0.05p = 0.05, the probability of needing more than kk trials is (1−p)k=(0.95)k(1-p)^k = (0.95)^k. So P(X>20)=(0.95)20P(X > 20) = (0.95)^{20} and P(X>10)=(0.95)10P(X > 10) = (0.95)^{10}. Therefore: P(X>20∣X>10)=(0.95)20(0.95)10=(0.95)20−10=(0.95)10P(X > 20 | X > 10) = \frac{(0.95)^{20}}{(0.95)^{10}} = (0.95)^{20-10} = (0.95)^{10} This makes intuitive sense: given that you've already tested 10 items without finding a defective one, you're essentially "starting over" and need 10 more consecutive non-defective items. Answer C, (0.95)10(0.95)^{10}, is correct. Answer A represents the conditional probability setup but isn't simplified. Answer B, (0.95)20(0.95)^{20}, gives the unconditional probability of testing more than 20 items. Answer D inverts the fraction, which would give an impossible probability greater than 1. Remember: geometric distributions are memoryless—past "failures" don't affect future probabilities. When you see "given that" with sequential testing, think conditional probability and the memoryless property.

Question 11

A search algorithm examines database entries until finding 4 entries matching specific criteria. Each entry has probability 16\frac{1}{6} of matching. What is the probability that fewer than 20 entries total will be examined?

  1. ∑k=419(k4)(16)4(56)k−4\sum_{k=4}^{19} \binom{k}{4} \left(\frac{1}{6}\right)^4 \left(\frac{5}{6}\right)^{k-4}
  2. ∑k=019(k+33)(16)4(56)k\sum_{k=0}^{19} \binom{k+3}{3} \left(\frac{1}{6}\right)^4 \left(\frac{5}{6}\right)^k
  3. ∑k=419(k−13)(16)4(56)k−4\sum_{k=4}^{19} \binom{k-1}{3} \left(\frac{1}{6}\right)^4 \left(\frac{5}{6}\right)^{k-4} (correct answer)
  4. ∑k=116(k+33)(16)4(56)k\sum_{k=1}^{16} \binom{k+3}{3} \left(\frac{1}{6}\right)^4 \left(\frac{5}{6}\right)^k
Explanation: When you encounter problems about searching until finding a specific number of successes, you're dealing with a negative binomial distribution. The key insight is recognizing what exactly is being counted and when the process stops. Here, the search stops immediately after finding the 4th matching entry. If fewer than 20 entries are examined total, this means the 4th success occurs on trial 4, 5, 6, ..., or 19. For the 4th success to occur on trial kk, you need exactly 3 successes in the first k−1k-1 trials, followed by a success on trial kk. The probability that the 4th success occurs on trial kk is (k−13)(16)4(56)k−4\binom{k-1}{3} \left(\frac{1}{6}\right)^4 \left(\frac{5}{6}\right)^{k-4}. The binomial coefficient (k−13)\binom{k-1}{3} counts ways to choose which 3 of the first k−1k-1 trials are successes. Summing from k=4k=4 to k=19k=19 gives the total probability, making C correct. A is wrong because (k4)\binom{k}{4} would mean choosing 4 successes from kk trials, but we need exactly the kk-th trial to be the 4th success. B uses the wrong binomial coefficient (k+33)\binom{k+3}{3} and sums over the wrong variable (number of failures rather than total trials). D has the same binomial coefficient error as B and an incorrect summation range. Strategy tip: In negative binomial problems, always identify what the "stopping trial" represents, then remember that for the rr-th success to occur on trial nn, you need (n−1r−1)\binom{n-1}{r-1} ways to arrange the first r−1r-1 successes.

Question 12

A quality inspector checks items sequentially until finding the 4th defective item. The probability of any item being defective is 0.150.15. Given that exactly 20 items were checked, what is the conditional probability that the 3rd defective item was found on the 15th check?

  1. (142)⋅(41)(193)\frac{\binom{14}{2} \cdot \binom{4}{1}}{\binom{19}{3}}
  2. (142)(193)\frac{\binom{14}{2}}{\binom{19}{3}} (correct answer)
  3. (153)(204)\frac{\binom{15}{3}}{\binom{20}{4}}
  4. 0.153⋅0.8512(193)⋅0.153⋅0.8516\frac{0.15^3 \cdot 0.85^{12}}{\binom{19}{3} \cdot 0.15^3 \cdot 0.85^{16}}
Explanation: This problem involves conditional probability in the context of sequential sampling, where you need to find the probability of a specific event given that another event has occurred. Given that exactly 20 items were checked and the 4th defective item was found on the 20th check, you need to find the probability that the 3rd defective item was found on the 15th check. Since the specific probabilities (0.15 defective, 0.85 non-defective) cancel out in conditional probability calculations involving only the positions of defective items, this becomes a purely combinatorial problem. The correct approach is to count favorable arrangements versus total possible arrangements. If the 3rd defective item is on the 15th check and the 4th is on the 20th check, then among the first 14 items, exactly 2 must be defective. This gives (142)\binom{14}{2} ways. The total number of ways to place the first 3 defective items among the first 19 positions is (193)\binom{19}{3}. Therefore, the answer is (142)(193)\frac{\binom{14}{2}}{\binom{19}{3}}, which is choice B. Choice A incorrectly includes (41)\binom{4}{1} in the numerator, suggesting confusion about which positions are being counted. Choice C uses (153)\binom{15}{3} and (204)\binom{20}{4}, which would be relevant for a different conditional probability question. Choice D unnecessarily includes the actual probabilities 0.15 and 0.85, which cancel out in conditional probability calculations when you're only concerned with the relative positions of events. Remember: in conditional probability problems involving positions or arrangements, focus on the combinatorial structure rather than getting distracted by specific probability values that often cancel out.

Question 13

A quality control inspector examines products until finding exactly 3 defective items. Each product has a probability of 0.080.08 of being defective. What is the probability that the inspector will examine exactly 25 products total?

  1. (242)⋅(0.08)3⋅(0.92)22\binom{24}{2} \cdot (0.08)^3 \cdot (0.92)^{22} (correct answer)
  2. (253)⋅(0.08)3⋅(0.92)22\binom{25}{3} \cdot (0.08)^3 \cdot (0.92)^{22}
  3. (243)⋅(0.08)3⋅(0.92)21\binom{24}{3} \cdot (0.08)^3 \cdot (0.92)^{21}
  4. (222)⋅(0.08)3⋅(0.92)22\binom{22}{2} \cdot (0.08)^3 \cdot (0.92)^{22}
Explanation: This is a negative binomial distribution. We need exactly 3 defective items in 25 trials, with the 25th being defective. This means 2 defective items in the first 24 trials, then 1 defective on trial 25. The probability is (242)⋅(0.08)2⋅(0.92)22⋅0.08=(242)⋅(0.08)3⋅(0.92)22\binom{24}{2} \cdot (0.08)^2 \cdot (0.92)^{22} \cdot 0.08 = \binom{24}{2} \cdot (0.08)^3 \cdot (0.92)^{22}. Choice B incorrectly treats this as a binomial. Choice C has wrong exponent on (0.92)(0.92). Choice D uses wrong binomial coefficient.

Question 14

In a geometric distribution with success probability p, if the probability of the first success occurring on the 3rd trial equals the probability of the first success occurring on the 5th trial, what is the value of p?

  1. p=14p = \frac{1}{4}
  2. p=13p = \frac{1}{3}
  3. p=12p = \frac{1}{2} (correct answer)
  4. p=23p = \frac{2}{3}
Explanation: When you encounter geometric distribution problems, remember that this distribution models the number of trials needed to achieve the first success, where each trial has the same probability pp of success. The probability that the first success occurs on trial kk is given by P(X=k)=(1−p)k−1⋅pP(X = k) = (1-p)^{k-1} \cdot p. This formula captures the idea that you need k−1k-1 failures followed by one success. Setting up the equation from the given condition:
  • Probability of first success on 3rd trial: P(X=3)=(1−p)2⋅pP(X = 3) = (1-p)^2 \cdot p
  • Probability of first success on 5th trial: P(X=5)=(1−p)4⋅pP(X = 5) = (1-p)^4 \cdot p
Since these probabilities are equal: (1−p)2⋅p=(1−p)4⋅p(1-p)^2 \cdot p = (1-p)^4 \cdot p Dividing both sides by pp (assuming p≠0p \neq 0): (1−p)2=(1−p)4(1-p)^2 = (1-p)^4 Dividing by (1−p)2(1-p)^2: 1=(1−p)21 = (1-p)^2 Taking the square root: 1−p=11-p = 1 or 1−p=−11-p = -1 Since pp must be a probability (between 0 and 1), we use 1−p=121-p = \frac{1}{2}, giving us p=12p = \frac{1}{2}. Answer A (p=14p = \frac{1}{4}) would make the 5th trial more likely than the 3rd. Answer B (p=13p = \frac{1}{3}) and Answer D (p=23p = \frac{2}{3}) also fail to satisfy the equality condition when substituted back into the original equation. Study tip: In geometric distribution problems involving equal probabilities at different trials, set up the probability equations and use algebraic manipulation. The key insight is recognizing when you can factor out common terms.

Question 15

In a manufacturing process, the probability of producing a defective item is 0.08. Using a negative binomial distribution, what is the probability that exactly 6 items must be produced to obtain the 3rd defective item?

  1. (52)×0.083×0.923\binom{5}{2} \times 0.08^3 \times 0.92^3 (correct answer)
  2. (63)×0.083×0.923\binom{6}{3} \times 0.08^3 \times 0.92^3
  3. (53)×0.083×0.922\binom{5}{3} \times 0.08^3 \times 0.92^2
  4. (62)×0.082×0.924\binom{6}{2} \times 0.08^2 \times 0.92^4
Explanation: For negative binomial distribution, to get the 3rd success on the 6th trial, we need exactly 2 successes in the first 5 trials, then a success on the 6th. This gives (52)×0.083×0.923\binom{5}{2} \times 0.08^3 \times 0.92^3. Choice B uses wrong binomial coefficient. Choice C has wrong power on 0.92. Choice D has wrong powers throughout.

Question 16

A quality control inspector tests electronic components until finding the first defective one. If each component has a 0.15 probability of being defective (independent of others), what is the probability that the inspector will test exactly 4 components before finding the first defective one?

  1. 0.853×0.150.85^3 \times 0.15 (correct answer)
  2. 0.854×0.150.85^4 \times 0.15
  3. (41)×0.853×0.15\binom{4}{1} \times 0.85^3 \times 0.15
  4. 0.853×0.1540.85^3 \times 0.15^4
Explanation: This is a geometric distribution problem. To test exactly 4 components means the first 3 are non-defective (probability 0.85 each) and the 4th is defective (probability 0.15). The probability is 0.853×0.150.85^3 \times 0.15. Choice B incorrectly includes an extra factor of 0.85. Choice C incorrectly applies binomial coefficient. Choice D has the wrong power on 0.15.

Question 17

A basketball player makes free throws with probability 0.7. If she continues shooting until she makes 4 successful shots, what is the expected number of shots she will need to take?

  1. 10.7+3=317\frac{1}{0.7} + 3 = \frac{31}{7}
  2. 40.3=403\frac{4}{0.3} = \frac{40}{3}
  3. 4×0.7=2.84 \times 0.7 = 2.8
  4. 40.7=407\frac{4}{0.7} = \frac{40}{7} (correct answer)
Explanation: When you encounter a problem about repeating an experiment until achieving a fixed number of successes, you're dealing with a negative binomial distribution. The key insight is that you need to find the expected number of trials to get a specified number of successes. For this type of problem, there's a fundamental formula: if you need rr successes and each trial has probability pp of success, then the expected number of trials is rp\frac{r}{p}. Here, the player needs r=4r = 4 successful shots with probability p=0.7p = 0.7 per shot, so the expected number of shots is 40.7=407\frac{4}{0.7} = \frac{40}{7}. Let's examine why the other answers miss the mark: Answer A attempts to use the formula for expected waiting time until the first success (10.7\frac{1}{0.7}) and then adds 3, but this incorrectly assumes the remaining shots after the first success follow a different pattern. Answer B uses 40.3\frac{4}{0.3} where 0.3 is the probability of failure. This reflects a common confusion between success and failure probabilities in the negative binomial formula. Answer C calculates 4×0.7=2.84 \times 0.7 = 2.8, which would give the expected number of successes in 4 trials, not the number of trials needed for 4 successes. Study tip: Remember the negative binomial expected value formula: E[trials]=successes neededprobability per trialE[\text{trials}] = \frac{\text{successes needed}}{\text{probability per trial}}. Always divide the target number of successes by the success probability, not the failure probability.

Question 18

A researcher conducts experiments until obtaining 5 successful results. Each experiment has a 0.4 probability of success. What is the probability that exactly 10 experiments are needed?

  1. (95)×(0.4)4×(0.6)5\binom{9}{5} \times (0.4)^4 \times (0.6)^5
  2. (105)×(0.4)5×(0.6)5\binom{10}{5} \times (0.4)^5 \times (0.6)^5
  3. (94)×(0.4)5×(0.6)5\binom{9}{4} \times (0.4)^5 \times (0.6)^5 (correct answer)
  4. (104)×(0.4)4×(0.6)6\binom{10}{4} \times (0.4)^4 \times (0.6)^6
Explanation: This problem involves the negative binomial distribution, which models the number of trials needed to achieve a fixed number of successes. When you see "experiments until obtaining X successes," think negative binomial rather than regular binomial distribution. For exactly 10 experiments to be needed for 5 successes, two conditions must be met: the 10th experiment must be a success (probability 0.4), and exactly 4 successes must occur in the first 9 experiments. This is the key insight that makes this a negative binomial scenario. The probability calculation breaks down as follows: We need (94)\binom{9}{4} ways to arrange 4 successes among the first 9 trials, times (0.4)4(0.4)^4 for those 4 successes, times (0.6)5(0.6)^5 for the 5 failures in those 9 trials, times 0.40.4 for the guaranteed success on trial 10. This gives us (94)×(0.4)5×(0.6)5\binom{9}{4} \times (0.4)^5 \times (0.6)^5, which is answer C. Answer A incorrectly uses (95)\binom{9}{5} and has the wrong power on the success probability. Answer B treats this as a regular binomial problem asking for exactly 5 successes in 10 trials, missing the constraint that the 10th trial must be a success. Answer D uses (104)\binom{10}{4}, which incorrectly includes the final trial in the combinatorial calculation, and has wrong probability powers. Remember: In "trials until X successes" problems, the final trial is always a success, so you only calculate combinations for the preceding trials. Look for this pattern on discrete math exams.

Question 19

A software tester runs test cases until encountering the first failure. If the probability of any test case failing is 0.25, what is the variance of the number of test cases that will be run?

  1. 0.750.25=3\frac{0.75}{0.25} = 3
  2. 0.25(1−0.25)2=49\frac{0.25}{(1-0.25)^2} = \frac{4}{9}
  3. 10.25=4\frac{1}{0.25} = 4
  4. 1−0.25(0.25)2=12\frac{1-0.25}{(0.25)^2} = 12 (correct answer)
Explanation: When you see a question about running trials until the first success or failure occurs, you're dealing with a geometric distribution. This distribution models the number of independent trials needed to achieve the first occurrence of an event with constant probability. For a geometric distribution where we count the trial on which the first success occurs, the variance formula is Var(X)=1−pp2\text{Var}(X) = \frac{1-p}{p^2}, where pp is the probability of success on each trial. In this problem, "success" means encountering a failure (which sounds contradictory but makes sense in context), so p=0.25p = 0.25. Applying the formula: Var(X)=1−0.25(0.25)2=0.750.0625=12\text{Var}(X) = \frac{1-0.25}{(0.25)^2} = \frac{0.75}{0.0625} = 12. This confirms answer D is correct. Looking at the wrong answers: Choice A gives 0.750.25=3\frac{0.75}{0.25} = 3, which is actually the formula for the mean of a geometric distribution (1−pp\frac{1-p}{p}), not the variance. Choice B shows 0.25(1−0.25)2=49\frac{0.25}{(1-0.25)^2} = \frac{4}{9}, which incorrectly puts pp in the numerator and (1−p)(1-p) in the denominator. Choice C gives 10.25=4\frac{1}{0.25} = 4, which is the standard mean formula 1p\frac{1}{p} for geometric distributions. Remember: for geometric distributions, always check whether the problem asks for mean or variance, and memorize both formulas. The variance formula has (1−p)(1-p) in the numerator and p2p^2 in the denominator—don't confuse it with the simpler mean formula.