Binomial Probability - Finite Mathematics

1. A new drug is effective in 60% of cases. The drug is administered to a group of 15 patients. What is the probability that the drug is effective for at least 8 patients but fewer than 11 patients?
   Answer: 0.597 Let $X$ be the number of patients for whom the drug is effective. This is a binomial distribution with $n=15$ and $p=0.6$. The question asks for the probability that $X$ is at least 8 but fewer than 11, which can be written as $P(8 \le X < 11)$. This means we need to find the sum of the probabilities for $X=8, X=9,$ and $X=10$. We calculate each probability: $P(X=8) = \binom{15}{8}(0.6)^8(0.4)^7 \approx 0.1771$. $P(X=9) = \binom{15}{9}(0.6)^9(0.4)^6 \approx 0.2066$. $P(X=10) = \binom{15}{10}(0.6)^{10}(0.4)^5 \approx 0.2131$. Summing these values gives $0.1771 + 0.2066 + 0.2131 = 0.5968 \approx 0.597$. Distractor B is $P(X=8)+P(X=9)$, an incomplete sum. Distractor D is $P(X \ge 8)$. Distractor A is $P(X=10)$, which is the probability for the upper end of the range, but not the whole range.
2. A student takes a 10-question multiple-choice quiz where each question has 4 options. The student decides to guess randomly on every question. What is the probability that the student gets exactly 3 questions correct, given that they get at least one question correct?
   Answer: 0.2652 Let $X$ be the number of correct answers. $X$ follows a binomial distribution $B(n, p)$ with $n=10$ trials and probability of success $p=1/4=0.25$. We want to find the conditional probability $P(X=3 | X \ge 1)$. The formula for conditional probability is $P(A|B) = P(A \cap B) / P(B)$. Here, $A$ is the event $X=3$ and $B$ is the event $X \ge 1$. The intersection $A \cap B$ is just $X=3$, since if $X=3$, then $X$ is also $\ge 1$. So we need to calculate $P(X=3) / P(X \ge 1)$. First, the numerator: $P(X=3) = \binom{10}{3}(0.25)^3(0.75)^7 \approx 0.2503$. Next, the denominator. It's easier to calculate $P(X \ge 1)$ using the complement rule: $P(X \ge 1) = 1 - P(X=0)$. $P(X=0) = \binom{10}{0}(0.25)^0(0.75)^{10} = (0.75)^{10} \approx 0.0563$. So, $P(X \ge 1) = 1 - 0.0563 = 0.9437$. Finally, the conditional probability is $0.2503 / 0.9437 \approx 0.2652$.
3. A quality control inspector examines a batch of 12 items. The probability that any given item is defective is 0.1. What is the probability that the batch contains at most two defective items?
   Answer: 0.8891 Let $X$ be the number of defective items in the batch. $X$ follows a binomial distribution with $n=12$ trials and probability of success (an item being defective) $p=0.1$. 'At most two' defective items means $X$ can be 0, 1, or 2. We need to calculate $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$. Calculate each probability: $P(X=0) = \binom{12}{0}(0.1)^0(0.9)^{12} = (0.9)^{12} \approx 0.2824$. $P(X=1) = \binom{12}{1}(0.1)^1(0.9)^{11} = 12(0.1)(0.9)^{11} \approx 0.3766$. $P(X=2) = \binom{12}{2}(0.1)^2(0.9)^{10} = 66(0.01)(0.9)^{10} \approx 0.2301$. Sum the probabilities: $P(X \le 2) \approx 0.2824 + 0.3766 + 0.2301 = 0.8891$.
4. Two archers, Alice and Bob, shoot at a target. Alice shoots 4 arrows with a probability of hitting the bullseye of 0.8 per shot. Bob shoots 6 arrows with a probability of hitting the bullseye of 0.5 per shot. What is the absolute difference between the probability that Alice hits at least 3 bullseyes and the probability that Bob hits at least 3 bullseyes?
   Answer: 0.163 First, calculate Alice's probability, $P(A \ge 3)$, with $n_A=4, p_A=0.8$. This is $P(A=3) + P(A=4) = \binom{4}{3}(0.8)^3(0.2)^1 + \binom{4}{4}(0.8)^4(0.2)^0 = 0.4096 + 0.4096 = 0.8192$. Second, calculate Bob's probability, $P(B \ge 3)$, with $n_B=6, p_B=0.5$. It is easier to use the complement: $P(B \ge 3) = 1 - [P(B=0) + P(B=1) + P(B=2)]$. $P(B=0) = \binom{6}{0}(0.5)^6 = 1/64$. $P(B=1) = \binom{6}{1}(0.5)^6 = 6/64$. $P(B=2) = \binom{6}{2}(0.5)^6 = 15/64$. The sum is $22/64$. So, $P(B \ge 3) = 1 - 22/64 = 42/64 = 0.65625$. The absolute difference is $|0.8192 - 0.65625| = 0.16295 \approx 0.163$. Distractor B is the difference in expected values. Distractor D is Alice's probability. Distractor C is $P(A=3)$.
5. A basketball player makes 70% of her free throws. In a practice session, she attempts 50 free throws. Let $X$ be the number of successful throws. What is the approximate probability that the number of successful throws she makes is within one standard deviation of the mean (inclusive)?
   Answer: 0.694 First, find the mean ($\mu$) and standard deviation ($\sigma$) of the binomial distribution. Here, $n=50$ and $p=0.7$. The mean is $\mu = np = 50(0.7) = 35$. The variance is $\sigma^2 = np(1-p) = 50(0.7)(0.3) = 10.5$. The standard deviation is $\sigma = \sqrt{10.5} \approx 3.24$. The interval within one standard deviation of the mean is $[\mu - \sigma, \mu + \sigma] = [35 - 3.24, 35 + 3.24] = [31.76, 38.24]$. We need to find the probability that $X$ falls in this interval. Since $X$ must be an integer, we sum the probabilities for $X = 32, 33, 34, 35, 36, 37, 38$. This sum is $P(32 \le X \le 38) = \sum_{k=32}^{38} \binom{50}{k}(0.7)^k(0.3)^{50-k} \approx 0.694$. Distractor A is the probability of the mean value only, $P(X=35)$. Distractor B is the sum for integers within 1 unit of the mean, $P(34 \le X \le 36)$. Distractor D is the approximate probability for being within two standard deviations.
6. A fair coin is tossed 20 times. Which of the following events has a probability equal to the probability of getting at most 7 heads?
   Answer: Getting at least 13 heads. For a binomial distribution with $p=0.5$, the probability distribution is symmetric about the mean $\mu = n/2$. In this case, $n=20$ and $p=0.5$, so the distribution is symmetric about $10$. For such a distribution, $P(X=k) = P(X=n-k)$. We want to find a probability equal to $P(X \le 7) = P(X=0) + P(X=1) + \dots + P(X=7)$. Using the symmetry property, this is equal to $P(X=20) + P(X=19) + \dots + P(X=13)$. This sum is precisely the probability of getting at least 13 heads, or $P(X \ge 13)$.
7. A software testing team finds bugs with probability 0.25 per test case, independently. They run test cases until finding 8 bugs, but will stop after 50 test cases even if fewer than 8 bugs are found. Given that they found exactly 8 bugs, what is the probability that they ran exactly 32 test cases?
   Answer: $\frac{\binom{31}{7}(0.25)^8(0.75)^{24}}{\sum_{k=8}^{50}\binom{k-1}{7}(0.25)^8(0.75)^{k-8} + \binom{50}{8}(0.25)^8(0.75)^{42}}$ This is a conditional probability problem involving a modified negative binomial distribution. When you see "stopping rules" combined with conditional probability, you need to carefully consider both the event space and the conditioning event. The correct answer is D because we need $P(\text{32 tests} | \text{exactly 8 bugs found})$. Using conditional probability: $\frac{P(\text{32 tests and 8 bugs})}{P(\text{exactly 8 bugs found})}$. For the numerator, if they ran exactly 32 tests and found 8 bugs, the 8th bug must have occurred on test 32 (otherwise they would have stopped earlier). This means 7 bugs in the first 31 tests, giving us $\binom{31}{7}(0.25)^8(0.75)^{24}$. For the denominator, "exactly 8 bugs found" can happen two ways: (1) they find the 8th bug on tests 8 through 49 and stop, or (2) they run all 50 tests and happen to find exactly 8 bugs. The first scenario contributes $\sum_{k=8}^{49}\binom{k-1}{7}(0.25)^8(0.75)^{k-8}$, and the second contributes $\binom{50}{8}(0.25)^8(0.75)^{42}$. Option A incorrectly treats this as a simple binomial problem, ignoring the stopping rule. Option B forgets the conditional probability denominator entirely. Option C misses the possibility that they could run all 50 tests and find exactly 8 bugs, only accounting for early stopping scenarios. Strategy tip: In stopping-time problems, always identify all ways the conditioning event can occur, especially when there's a maximum limit that could be reached.
8. A city's bus service claims that buses arrive on time with a probability of $p=0.9$. An observer monitors 10 randomly selected bus arrivals during a single morning's rush hour to test this claim. Let $X$ be the number of on-time arrivals. Calculating $P(X \ge 8)$ using the binomial model relies on several assumptions. Which of the following represents the most likely violation of the binomial model's assumptions in this specific context?
   Answer: The outcomes of the trials are not independent or the probability of success is not constant. The binomial distribution relies on four key assumptions (BINS): Binary outcomes, Independent trials, a fixed Number of trials, and the same probability of Success for each trial. In this scenario: (A) is incorrect because the binomial model is exact for any fixed integer $n \ge 1$. (C) is incorrect because the outcome is defined as binary (on-time or not on-time). (D) is incorrect because the expected value being an integer has no bearing on the validity of the model. The most likely violation is (B). During rush hour, a delay in one bus due to traffic could have a cascading effect on subsequent buses, violating the independence of trials. Furthermore, traffic conditions can worsen over the rush hour period, meaning the probability of a bus being on time might not be constant for all 10 observations.
9. A manufacturer must ship electronic components in batches. The probability that a component is defective is 2%. What is the minimum batch size required to ensure that the probability of finding at least one defective component in the batch is greater than 90%?
   Answer: 114 Let $n$ be the batch size and $p=0.02$ be the probability of a component being defective. The probability of finding at least one defective component is $P(X \ge 1) = 1 - P(X=0)$. We want this to be greater than 0.90. So, $1 - P(X=0) > 0.90$. The probability of zero defective components is $P(X=0) = \binom{n}{0}(0.02)^0(0.98)^n = (0.98)^n$. The inequality becomes $1 - (0.98)^n > 0.90$, which simplifies to $0.10 > (0.98)^n$. To solve for $n$, we take the logarithm of both sides: $\log(0.10) > n \log(0.98)$. Since $\log(0.98)$ is negative, we must reverse the inequality sign when dividing: $n > \frac{\log(0.10)}{\log(0.98)} \approx 113.97$. Since $n$ must be an integer, the minimum batch size is 114. Distractor B is a rounding error. Distractor A results from incorrectly setting up the problem as $np > 0.9$. Distractor D is the result of solving for a probability greater than 99%.
10. A basketball player makes free throws with probability 0.72. During practice, she attempts sets of 10 free throws. If she makes at least 8 out of 10 in a set, she stops practicing. Otherwise, she attempts another set. What is the probability that she stops after exactly 3 sets?
    Answer: $[P(X \leq 7)]^2 \cdot P(X \geq 8) \text{ where } X \sim \text{Binomial}(10, 0.72)$ For her to stop after exactly 3 sets, she must fail to make 8+ in the first two sets (probability P(X ≤ 7) each) and then succeed in the third set (probability P(X ≥ 8)). The probability is [P(X ≤ 7)]² · P(X ≥ 8). Note that P(X ≤ 7) = 1 - P(X ≥ 8), so choice A is mathematically equivalent to choice C, but choice C more directly expresses the required conditions. Choice B incorrectly uses a binomial coefficient as if we're choosing which set succeeds, but the success must be on the 3rd set specifically. Choice D unnecessarily introduces binomial coefficients for a deterministic sequence.
11. In a manufacturing process, each item has a 0.12 probability of being defective. A batch is considered acceptable if it contains at most 2 defective items out of 15. However, due to a calibration error, the actual defect rate is 0.18. What is the difference between the probability of acceptance calculated using the assumed rate versus the actual rate?
    Answer: $\sum_{k=0}^{2}\binom{15}{k}[(0.12)^k(0.88)^{15-k} - (0.18)^k(0.82)^{15-k}] \approx 0.1842$ We need P(X ≤ 2) with p = 0.12 minus P(X ≤ 2) with p = 0.18, where X ~ Binomial(15, p). This equals the sum from k=0 to 2 of the difference between the individual binomial probabilities. Choice B incorrectly tries to use the differences (0.18-0.12) and (0.88-0.82) as if they were new parameters. Choice C only considers k=2 instead of k=0,1,2. Choice D uses 0.06 as a probability parameter, which doesn't correspond to either scenario.
12. A factory produces light bulbs with a 10% defect rate. A quality inspector takes a random sample of 18 bulbs. What is the probability that the sample contains the most likely number of defective bulbs?
    Answer: 0.300 This question requires two steps: first find the most likely number of successes (the mode), then calculate its probability. For a binomial distribution with $n=18$ and $p=0.1$, the mode is the integer part of $(n+1)p$. Mode $= \lfloor (18+1)(0.1) \rfloor = \lfloor 1.9 \rfloor = 1$. So, the most likely number of defective bulbs is 1. Now, we calculate the probability of getting exactly 1 defective bulb: $P(X=1) = \binom{18}{1}(0.1)^1(0.9)^{17} = 18(0.1)(0.16677) \approx 0.300$. Distractor B is the probability of 2 defective bulbs, which is a common mistake as the mean ($np=1.8$) is closer to 2 than 1. Distractor A is $P(X=0)$. Distractor D is a simple fraction $1/3$, a plausible guess.
13. For a binomial experiment with $n=5$ trials, the probability of obtaining exactly 3 successes is exactly 10 times the probability of obtaining exactly 4 successes. Assuming the probability of success $p$ is not zero, what is the value of $p$?
    Answer: $1/6$ Let $p$ be the probability of success and $1-p$ be the probability of failure. The probability of $k$ successes in $n$ trials is given by the binomial formula $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$. We are given $n=5$ and the condition $P(X=3) = 10 \cdot P(X=4)$. Let's write out the expressions for these probabilities: $P(X=3) = \binom{5}{3}p^3(1-p)^{5-3} = 10p^3(1-p)^2$. $P(X=4) = \binom{5}{4}p^4(1-p)^{5-4} = 5p^4(1-p)^1$. Now, set up the equation based on the given condition: $10p^3(1-p)^2 = 10 \cdot [5p^4(1-p)]$. $10p^3(1-p)^2 = 50p^4(1-p)$. Since we are told $p \ne 0$, we know $p^3 \ne 0$. Also, if $p=1$, then $P(X=3)=0$ and $P(X=4)=0$, which satisfies the equation, but we assume a non-trivial case. If $p \ne 1$, we can divide both sides by $10p^3(1-p)$: $(1-p) = 5p$. $1 = 6p$. $p = 1/6$.
14. The number of successful trials in a binomial experiment has a mean of 4 and a variance of 2.4. What is the probability of observing exactly 3 successful trials?
    Answer: 0.215 For a binomial distribution, the mean is $\mu = np$ and the variance is $\sigma^2 = np(1-p)$. We are given $\mu=4$ and $\sigma^2=2.4$. We can set up a system of equations: (1) $np = 4$ and (2) $np(1-p) = 2.4$. Substitute (1) into (2) to get $4(1-p) = 2.4$. Solving for $p$, we find $1-p = 0.6$, so $p=0.4$. Substitute $p=0.4$ back into (1) to get $n(0.4) = 4$, which gives $n=10$. Now we need to calculate $P(X=3)$ for a binomial distribution with $n=10$ and $p=0.4$. The formula is $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$. So, $P(X=3) = \binom{10}{3}(0.4)^3(0.6)^7 \approx 120(0.064)(0.02799) \approx 0.215$. Distractor D is $P(X=4)$, the probability of the mean outcome. Distractor A is the result of incorrectly using $p=0.6$. Distractor B is $P(X=2)$.
15. A telemarketer makes 20 calls. The probability of a sale on any given call is 0.1. For each sale, the company profits $100. For each unsuccessful call, the company loses $10. What is the probability that the company makes a positive profit from the 20 calls?
    Answer: 0.608 Let $k$ be the number of sales (successes) in $n=20$ calls. The number of unsuccessful calls is $20-k$. The total profit is $Profit = 100k - 10(20-k)$. We want the profit to be positive: $100k - 200 + 10k > 0 \implies 110k > 200 \implies k > 200/110 \approx 1.818$. Since $k$ must be an integer, the company makes a profit if $k \ge 2$. We need to calculate $P(X \ge 2)$ for a binomial distribution with $n=20$ and $p=0.1$. It is easiest to use the complement: $P(X \ge 2) = 1 - [P(X=0) + P(X=1)]$. $P(X=0) = \binom{20}{0}(0.1)^0(0.9)^{20} \approx 0.1216$. $P(X=1) = \binom{20}{1}(0.1)^1(0.9)^{19} \approx 0.2702$. So, $P(X \ge 2) = 1 - (0.1216 + 0.2702) = 1 - 0.3918 = 0.6082$. Distractor C is approximately 0.608. Distractor B is the probability of not making a profit, $P(X \le 1)$. Distractor A is $P(X=2)$. Distractor D is $P(X \ge 1)$.
16. Player A flips a fair coin 5 times and wins if they get more than 3 heads. Player B rolls a fair 6-sided die 4 times and wins if they roll at least one '6'. Which statement accurately compares their probabilities of winning?
    Answer: Player B has a higher probability of winning by approximately 0.32. When comparing probabilities in games of chance, you need to calculate each player's exact probability of winning using binomial distributions and complement rules. For Player A to win, they need more than 3 heads in 5 flips, meaning exactly 4 or 5 heads. Using the binomial formula with $n=5$ and $p=0.5$: - P(4 heads) = $\binom{5}{4}(0.5)^5 = 5 \times \frac{1}{32} = \frac{5}{32}$ - P(5 heads) = $\binom{5}{5}(0.5)^5 = 1 \times \frac{1}{32} = \frac{1}{32}$ So P(Player A wins) = $\frac{5}{32} + \frac{1}{32} = \frac{6}{32} = 0.1875$ For Player B to win, they need at least one '6' in 4 rolls. It's easier to use the complement: P(at least one 6) = 1 - P(no 6s). The probability of not rolling a 6 on any single roll is $\frac{5}{6}$, so: P(Player B wins) = $1 - \left(\frac{5}{6}\right)^4 = 1 - \frac{625}{1296} = \frac{671}{1296} \approx 0.518$ Player B's advantage is $0.518 - 0.1875 = 0.3305 \approx 0.32$, confirming answer C. Answer A incorrectly reverses who has the advantage. Answer B gets the direction wrong and uses an incorrect difference. Answer D has the wrong direction and magnitude—likely from calculation errors in the binomial probabilities. Study tip: When calculating "at least one" probabilities, always consider using the complement rule (1 minus "none")—it's usually much simpler than adding multiple probability terms.
17. A manufacturer of microchips finds that 5% of its chips are defective. A quality control process involves testing a random sample of 50 chips from a large production batch. Let $X$ be the number of defective chips in the sample. What is the probability that the number of defective chips is within one standard deviation of the mean number of defective chips?
    Answer: 0.8195 The number of defective chips $X$ follows a binomial distribution $B(n, p)$ with $n=50$ and $p=0.05$. First, we calculate the mean (expected value) $\mu$ and standard deviation $\sigma$. Mean: $\mu = np = 50(0.05) = 2.5$. Variance: $\sigma^2 = np(1-p) = 50(0.05)(0.95) = 2.375$. Standard deviation: $\sigma = \sqrt{2.375} \approx 1.541$. We need to find the probability that $X$ is within one standard deviation of the mean, which is the interval $[\mu - \sigma, \mu + \sigma]$. This interval is $[2.5 - 1.541, 2.5 + 1.541]$, or approximately $[0.959, 4.041]$. The integer values for $X$ (number of defective chips) in this interval are 1, 2, 3, and 4. We need to calculate $P(X=1) + P(X=2) + P(X=3) + P(X=4)$. $P(X=1) = \binom{50}{1}(0.05)^1(0.95)^{49} \approx 0.2025$ $P(X=2) = \binom{50}{2}(0.05)^2(0.95)^{48} \approx 0.2611$ $P(X=3) = \binom{50}{3}(0.05)^3(0.95)^{47} \approx 0.2199$ $P(X=4) = \binom{50}{4}(0.05)^4(0.95)^{46} \approx 0.1360$ The sum is $0.2025 + 0.2611 + 0.2199 + 0.1360 = 0.8195$.
18. An archer has a 0.8 probability of hitting a target with each arrow. Let $A$ be the event that the archer's first hit occurs on the third shot. Let $B$ be the event that in a fixed set of 3 shots, the archer hits the target exactly once. What is the value of the absolute difference $|P(A) - P(B)|$?
    Answer: 0.064 This question requires calculating two different probabilities. The probability of hitting is $p=0.8$, and the probability of missing is $q=0.2$. Event A involves a geometric distribution, as we are waiting for the first success. The first hit occurring on the third shot means the sequence of outcomes must be Miss, Miss, Hit. The probability of this specific sequence is $P(A) = q \cdot q \cdot p = (0.2)^2(0.8) = 0.04 \cdot 0.8 = 0.032$. Event B involves a binomial distribution with $n=3$ trials and $p=0.8$. We want the probability of exactly one success ($k=1$). The formula is $P(B) = P(X=1) = \binom{3}{1}p^1q^{3-1} = 3(0.8)^1(0.2)^2 = 3(0.8)(0.04) = 2.4(0.04) = 0.096$. The absolute difference is $|P(A) - P(B)| = |0.032 - 0.096| = |-0.064| = 0.064$.
19. For a binomial distribution $B(n, p)$, the distribution is symmetric when $p=0.5$. If such a symmetric binomial distribution has two distinct modes (i.e., it is bimodal), which of the following statements about the number of trials $n$ must be true?
    Answer: $n$ must be an odd number. When you encounter questions about binomial distribution modes, you need to understand how the shape and peaks of the distribution relate to the parameters $n$ and $p$. For a binomial distribution $B(n, p)$, a mode occurs at any value $k$ where the probability $P(X = k)$ is maximized. The key insight is that when $p = 0.5$, the distribution is symmetric around the mean $\mu = np = 0.5n$. For the distribution to be bimodal (having two distinct peaks), these peaks must be equidistant from the center. When $p = 0.5$ and $n$ is odd, the mean $0.5n$ falls exactly halfway between two integers. For example, if $n = 5$, then $\mu = 2.5$, and the distribution has modes at both $k = 2$ and $k = 3$. This creates the bimodal pattern because no single integer value coincides with the mean. Looking at the answer choices: Choice B is incorrect because when $n$ is even, $0.5n$ is an integer, making the distribution unimodal with a single peak at that integer value. Choice C is wrong since bimodality only occurs for odd $n$ values when $p = 0.5$. Choice D incorrectly focuses on whether the mean is an integer – this is actually the determining factor, but it's not the direct constraint on $n$. Therefore, A is correct: $n$ must be odd. Study tip: Remember that for symmetric binomial distributions, bimodality occurs when the mean falls between two integers, which happens when $n$ is odd and $p = 0.5$.
20. A basketball player makes 70% of her free throws. What is the minimum number of free throws she must attempt to make the probability of hitting at least one shot greater than 0.999?
    Answer: 6 Let $n$ be the number of free throws attempted. Let $X$ be the number of successful shots. The probability of success on a single trial is $p=0.7$. The probability of failure is $1-p=0.3$. The event 'at least one success' is the complement of the event 'zero successes'. $P(X \ge 1) = 1 - P(X=0)$. We are given the condition $P(X \ge 1) > 0.999$. Substituting the complement formula: $1 - P(X=0) > 0.999$. $P(X=0)$ is the probability of failing on all $n$ attempts, which is $(0.3)^n$. So, the inequality is $1 - (0.3)^n > 0.999$. Rearranging the inequality: $- (0.3)^n > -0.001$ $(0.3)^n < 0.001$ To solve for $n$, we take the logarithm of both sides. Since $\log(x)$ is an increasing function for $x>0$, the inequality direction is preserved if the base is greater than 1. $\log((0.3)^n) < \log(0.001)$ $n \log(0.3) < \log(0.001)$ Since $\log(0.3)$ is a negative number (as $0.3 < 1$), we must reverse the inequality sign when we divide by it. $n > \frac{\log(0.001)}{\log(0.3)} = \frac{-3}{-0.5228} \approx 5.73$. Since $n$ must be an integer, the minimum number of attempts is 6.

Start this quiz