Discrete Math Quiz: Binomial Distribution

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Binomial DistributionQuestion 1 of 19

A manufacturing process produces components where 8% are defective. Components are tested in batches of 25. If a batch contains 3 or more defective components, it is rejected. What is the probability that a batch is rejected?

1 - \left[\binom{25}{0}(0.92)^{25} + \binom{25}{1}(0.08)(0.92)^{24} + \binom{25}{2}(0.08)^2(0.92)^{23}\right]

\binom{25}{3}(0.08)^3(0.92)^{22}

\sum_{k=3}^{22} \binom{25}{k}(0.08)^k(0.92)^{25-k}

1 - \left[\binom{25}{0}(0.08)^0(0.92)^{25} + \binom{25}{1}(0.08)^1(0.92)^{24} + \binom{25}{2}(0.08)^2(0.92)^{23}\right]

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Discrete Math Quiz

Discrete Math Quiz: Binomial Distribution

Practice Binomial Distribution 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 Binomial Distribution, 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

$1 - \left[\binom{25}{0}(0.92)^{25} + \binom{25}{1}(0.08)(0.92)^{24} + \binom{25}{2}(0.08)^2(0.92)^{23}\right]$
$\binom{25}{3}(0.08)^3(0.92)^{22}$
$\sum_{k=3}^{22} \binom{25}{k}(0.08)^k(0.92)^{25-k}$
$1 - \left[\binom{25}{0}(0.08)^0(0.92)^{25} + \binom{25}{1}(0.08)^1(0.92)^{24} + \binom{25}{2}(0.08)^2(0.92)^{23}\right]$ (correct answer)

Explanation: When you encounter a problem about defective items in batches, you're dealing with a binomial probability distribution. Each component has a fixed probability of being defective (8%), and you're looking at a specific number of trials (25 components). The key insight is recognizing that finding "3 or more defective" is easiest calculated using the complement rule: P(≥3 defective) = 1 - P(≤2 defective). This means you calculate the probability of getting exactly 0, 1, or 2 defective components, then subtract from 1. Using the binomial probability formula

P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}

, where n = 25, p = 0.08, you need:

P(0 defective) = $\binom{25}{0}(0.08)^0(0.92)^{25}$
P(1 defective) = $\binom{25}{1}(0.08)^1(0.92)^{24}$
P(2 defective) = $\binom{25}{2}(0.08)^2(0.92)^{23}$

Answer D correctly applies this complement approach with proper binomial formula notation. Answer A has the right structure but incorrectly writes the first two terms without proper exponents on 0.08. Answer B only calculates P(exactly 3 defective), missing all cases from 4 to 25 defective components. Answer C uses direct summation from k=3 to k=22, but stops at 22 instead of going to 25, missing the cases with 23, 24, or 25 defective components. Study tip: For "at least X" or "X or more" problems in binomial distributions, always consider whether the complement approach (1 minus the opposite) will be computationally easier than direct calculation.

Question 2

A pharmaceutical trial tests a new drug on patients, where each patient independently has a 0.65 probability of showing improvement. The trial will continue until exactly 20 patients show improvement. What is the probability that exactly 30 patients will need to be tested?

$\binom{29}{19}(0.65)^{19}(0.35)^{10}(0.65)$
$\binom{30}{20}(0.65)^{20}(0.35)^{10}$
$\binom{29}{20}(0.65)^{20}(0.35)^9$
$\binom{29}{19}(0.65)^{20}(0.35)^{10}$ (correct answer)

Explanation: When you encounter a problem about continuing trials until a fixed number of successes occurs, you're dealing with a negative binomial distribution. This differs from regular binomial problems because the number of trials is variable, not fixed. Here's the key insight: if exactly 30 patients need to be tested to get 20 improvements, then the 30th patient must show improvement (that's what stops the trial), and among the first 29 patients, exactly 19 must have shown improvement. The probability breaks down into two parts:

The probability that exactly 19 of the first 29 patients improve: $\binom{29}{19}(0.65)^{19}(0.35)^{10}$
The probability that the 30th patient improves: $0.65$

Multiplying these gives:

\binom{29}{19}(0.65)^{20}(0.35)^{10}

, which is answer D. A incorrectly separates the final success probability as

(0.65)

at the end rather than incorporating it into the exponent. B uses

\binom{30}{20}

, treating this as a standard binomial problem where any 20 of 30 patients could be the successes—but this ignores that the 30th patient must be a success to stop the trial. C has the wrong binomial coefficient

\binom{29}{20}

(impossible since we need exactly 19 successes in 29 trials) and uses

(0.35)^9

instead of

(0.35)^{10}

. Study tip: In negative binomial problems, always remember that the final trial must be a success. Work backwards: if you need exactly

n

trials to get

r

successes, then exactly

r-1

successes must occur in the first

n-1

trials.

Question 3

In a manufacturing process, the probability that any single item is defective is 0.08. A batch is considered acceptable if it contains at most 2 defective items out of 20. What is the probability that a randomly selected batch is acceptable?

$\sum_{k=0}^{2} \binom{20}{k}(0.08)^k(0.92)^{20-k}$ (correct answer)
$\sum_{k=0}^{18} \binom{20}{k}(0.08)^k(0.92)^{20-k}$
$\binom{20}{2}(0.08)^2(0.92)^{18}$
$1 - \sum_{k=3}^{20} \binom{20}{k}(0.08)^k(0.92)^{20-k}$

Explanation: We need P(X ≤ 2) where X ~ Binomial(20, 0.08). This equals the sum of P(X=0), P(X=1), and P(X=2). Choice A correctly sums from k=0 to k=2. Choice B sums too many terms (up to 18 defective items). Choice C only calculates P(X=2). Choice D is mathematically equivalent to A but unnecessarily complex.

Question 4

A quality control inspector tests electronic components, where each component has a 0.85 probability of passing inspection independently. If the inspector needs to find exactly 12 components that pass out of the first 15 components tested, what is the probability of this occurring?

$\binom{15}{12}(0.85)^{12}(0.15)^3$ (correct answer)
$\binom{12}{15}(0.85)^{12}(0.15)^3$
$\binom{15}{12}(0.85)^{15}(0.15)^0$
$\binom{15}{3}(0.85)^{12}(0.15)^3$

Explanation: This is a binomial distribution with n=15 trials, k=12 successes, and p=0.85. The formula is

P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}

. Choice B has the binomial coefficient backwards. Choice C uses the wrong exponents (should be 3 failures, not 0). Choice D uses

\binom{15}{3}

instead of

\binom{15}{12}

(though these are equal, the standard form uses k=12).

Question 5

A basketball player makes free throws with probability 0.6. During practice, she will shoot free throws until she makes exactly 8 successful shots. What type of probability distribution best models the total number of free throws she will attempt?

Binomial distribution with parameters n and p = 0.6
Negative binomial distribution with parameters r = 8 and p = 0.6 (correct answer)
Geometric distribution with parameter p = 0.6
Poisson distribution with parameter λ = 8

Explanation: This describes a negative binomial distribution because we're counting the number of trials needed to achieve a fixed number of successes (r=8). Choice A is wrong because binomial distribution has a fixed number of trials. Choice C is wrong because geometric distribution counts trials until the first success. Choice D is wrong because Poisson distribution models rare events over time/space intervals.

Question 6

In a quality control process, items are inspected with each item having a 0.15 probability of being defective. Inspectors want to determine the minimum sample size n such that P(X ≥ 1) ≥ 0.90, where X is the number of defective items found.

n = 14
n = 15
n = 16 (correct answer)
n = 17

Explanation: We need P(X ≥ 1) ≥ 0.90, which means 1 - P(X = 0) ≥ 0.90, so P(X = 0) ≤ 0.10. Since P(X = 0) = (0.85)ⁿ, we need (0.85)ⁿ ≤ 0.10. Taking logarithms: n log(0.85) ≤ log(0.10), so n ≥ log(0.10)/log(0.85) ≈ 15.3. Since n must be an integer, n = 16. Choice A and B are too small. Choice D is larger than necessary.

Question 7

A pharmaceutical company claims their new drug has a 75% success rate. To test this claim, researchers will administer the drug to 25 patients. If the company's claim is true, what is the expected number of successful treatments and the standard deviation?

Expected value: 18.75, Standard deviation: 2.17 (correct answer)
Expected value: 18.75, Standard deviation: 4.69
Expected value: 6.25, Standard deviation: 2.17
Expected value: 12.5, Standard deviation: 3.54

Explanation: For a binomial distribution with n=25 and p=0.75: Expected value = np = 25(0.75) = 18.75. Variance = np(1-p) = 25(0.75)(0.25) = 4.6875. Standard deviation = √4.6875 ≈ 2.17. Choice B uses variance instead of standard deviation. Choice C uses the expected number of failures (6.25). Choice D uses incorrect calculations.

Question 8

A server processes requests independently, where each request has probability $0.3$ of requiring additional resources. The server can handle at most $2$ resource-intensive requests simultaneously. If $8$ requests arrive, what is the probability that the server becomes overloaded (more than $2$ requests require additional resources)?

$\binom{8}{3}(0.3)^3(0.7)^5 + \binom{8}{4}(0.3)^4(0.7)^4$
$\sum_{k=3}^{8} \binom{8}{k}(0.7)^k(0.3)^{8-k}$
$1 - (0.7)^8 - \binom{8}{1}(0.3)(0.7)^7 - \binom{8}{2}(0.3)^2(0.7)^6$
$1 - \sum_{k=0}^{2} \binom{8}{k}(0.3)^k(0.7)^{8-k}$ (correct answer)

Explanation: When you encounter probability questions about "at most" or "more than" a certain number of events, think about whether it's easier to calculate directly or use the complement rule. Here, finding the probability of overload (more than 2 resource-intensive requests) is much simpler using complements. This is a binomial probability problem where

n=8

requests, each with probability

p=0.3

of requiring resources. The server becomes overloaded when 3 or more requests need resources, so we want

P(X \geq 3)

where

X

follows a binomial distribution. Using the complement rule:

P(X \geq 3) = 1 - P(X \leq 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]

. This equals

1 - \sum_{k=0}^{2} \binom{8}{k}(0.3)^k(0.7)^{8-k}

, which is answer D. Answer A only includes the probabilities for exactly 3 and exactly 4 resource-intensive requests, missing cases where 5, 6, 7, or 8 requests need resources. Answer B has the right structure for summing from 3 to 8, but incorrectly swaps the probabilities—it uses

(0.7)^k

when it should be

(0.3)^k

. Answer C writes out the complement formula correctly in expanded form, making it equivalent to answer D, but it's not one of the standard forms typically expected. Strategy tip: For binomial "at least" or "more than" problems, always check if the complement approach (calculating "at most" the opposite boundary) involves fewer terms. It's usually faster and less error-prone than summing multiple probability terms directly.

Question 9

A manufacturing process produces items where each item independently has probability $p$ of meeting quality standards. The process continues until either $5$ acceptable items are produced or $3$ unacceptable items are produced, whichever occurs first. If $p = 0.6$ , what is the probability that the process stops with exactly $7$ total items produced?

$\binom{6}{4}(0.6)^5(0.4)^2 + \binom{6}{3}(0.6)^4(0.4)^3$ (correct answer)
$\binom{6}{2}(0.6)^5(0.4)^2 + \binom{6}{4}(0.6)^3(0.4)^3$
$\binom{6}{2}(0.6)^5(0.4)^2 + \binom{6}{3}(0.6)^4(0.4)^3$
$\binom{6}{4}(0.6)^5(0.4)^2 + \binom{6}{2}(0.6)^4(0.4)^3$

Explanation: For exactly 7 items, the process stops on the 7th item due to either: (1) getting the 5th acceptable item (with exactly 2 unacceptable among first 6), or (2) getting the 3rd unacceptable item (with exactly 4 acceptable among first 6). Case 1: Among the first 6 items, we need exactly 4 acceptable items, so

\binom{6}{4}(0.6)^4(0.4)^2

, then multiply by 0.6 for the 7th acceptable item:

\binom{6}{4}(0.6)^5(0.4)^2

. Case 2: Among the first 6 items, we need exactly 3 unacceptable items, so

\binom{6}{3}(0.6)^4(0.4)^3

. The other choices use incorrect binomial coefficients for the two cases.

Question 10

A quality control inspector tests electronic components, where each component has a $0.15$ probability of being defective, independent of other components. The inspector needs to find at least $2$ defective components among the next $8$ components tested to trigger a full production line review. What is the probability that a review will be triggered?

$1 - (0.85)^8 - 8(0.15)(0.85)^7$
$\binom{8}{2}(0.15)^2(0.85)^6$
$1 - (0.85)^8 - \binom{8}{1}(0.15)(0.85)^7$ (correct answer)
$\sum_{k=2}^{8} \binom{8}{k}(0.85)^k(0.15)^{8-k}$

Explanation: We need P(X ≥ 2) where X ~ Binomial(8, 0.15). Using the complement: P(X ≥ 2) = 1 - P(X < 2) = 1 - P(X = 0) - P(X = 1) = 1 - (0.85)^8 - C(8,1)(0.15)(0.85)^7. Choice A omits the binomial coefficient for k=1. Choice B only gives P(X = 2). Choice D incorrectly swaps the probabilities in the binomial terms.

Question 11

A quality inspector tests circuit boards in batches of $15$ . Each board independently has a $0.08$ probability of being defective. Using a normal approximation to the binomial distribution, what is the approximate probability that a batch contains more than $2$ defective boards?

$P(Z \geq \frac{3 - 1.2}{\sqrt{1.104}})$ where $Z$ is standard normal
$P(Z > \frac{2 - 1.2}{\sqrt{1.104}})$ where $Z$ is standard normal
$1 - P(Z \leq \frac{2.5 - 1.2}{\sqrt{1.104}})$ where $Z$ is standard normal
$P(Z > \frac{2.5 - 1.2}{\sqrt{1.104}})$ where $Z$ is standard normal (correct answer)

Explanation: When you encounter a problem asking for a normal approximation to a binomial distribution, you need to identify the binomial parameters, apply continuity correction, and standardize properly. This is a binomial situation with

n = 15

trials and

p = 0.08

probability of success (defective board). The mean is

\mu = np = 15 \times 0.08 = 1.2

and variance is

\sigma^2 = np(1-p) = 15 \times 0.08 \times 0.92 = 1.104

, so

\sigma = \sqrt{1.104}

. You want

P(X > 2)

where

X

is the number of defective boards. Since you're approximating a discrete distribution with a continuous one, you must apply continuity correction. "More than 2" becomes "greater than 2.5" in the continuous approximation. So you need

P(X > 2.5)

. Standardizing:

P(Z > \frac{2.5 - 1.2}{\sqrt{1.104}})

where

Z

is standard normal. Answer D is correct because it properly applies continuity correction (2.5 instead of 2) and uses the correct inequality direction. Answer A is wrong because it uses 3 instead of 2.5 for the continuity correction and uses

\geq

instead of

>

. Answer B is wrong because it fails to apply continuity correction, using 2 instead of 2.5. Answer C is mathematically equivalent to D but unnecessarily complicated, using the complement

1 - P(Z \leq 2.5...)

instead of the direct

P(Z > 2.5...)

. Key strategy: Always apply continuity correction when using normal approximation to binomial. For "more than

k

", use

k + 0.5

in your calculation.

Question 12

A basketball player makes free throws with probability $0.7$ . She will keep shooting until she either makes $3$ shots or misses $3$ shots, whichever comes first. What is the probability that she stops because she made $3$ shots?

$\sum_{k=0}^{2} \binom{k+2}{2}(0.7)^3(0.3)^k$ (correct answer)
$\sum_{k=0}^{2} \binom{k+2}{k}(0.7)^3(0.3)^k$
$\binom{5}{2}(0.7)^3(0.3)^2 + \binom{4}{1}(0.7)^3(0.3)^1 + (0.7)^3$
$\binom{4}{2}(0.7)^3(0.3)^2 + \binom{3}{1}(0.7)^3(0.3)^1 + (0.7)^3$

Explanation: She stops with 3 makes if she gets her 3rd make before her 3rd miss. This means she can have 0, 1, or 2 misses before getting 3 makes. For k misses, she needs k+3 total shots, with the last shot being a make and exactly k misses among the first k+2 shots. This gives C(k+2,k)(0.7)^3(0.3)^k = C(k+2,2)(0.7)^3(0.3)^k. Choice B has the wrong binomial coefficient form. Choices C and D attempt direct enumeration but use incorrect shot counts.

Question 13

An online quiz has $12$ multiple-choice questions, each with $4$ options. A student who knows the answers to exactly $8$ questions guesses randomly on the remaining $4$ questions. What is the probability that the student gets exactly $10$ questions correct?

$\binom{12}{10}(\frac{3}{4})^{10}(\frac{1}{4})^{2}$
$\binom{4}{2}(\frac{1}{4})^{2}(\frac{3}{4})^{2}$ (correct answer)
$\binom{8}{8}\binom{4}{2}(\frac{1}{4})^{2}(\frac{3}{4})^{2}$
$\binom{4}{2}(\frac{1}{4})^{2}(\frac{3}{4})^{2} + \binom{8}{2}(\frac{1}{4})^{6}$

Explanation: The student knows 8 answers (gets all 8 correct) and must get exactly 2 out of 4 guessed questions correct. For the 4 guessed questions with probability 1/4 each, this is a binomial: P(exactly 2 correct) = C(4,2)(1/4)^2(3/4)^2. Choice A treats all 12 questions as having probability 3/4, ignoring that 8 are known. Choice C unnecessarily includes C(8,8)=1. Choice D incorrectly adds terms and has wrong probabilities.

Question 14

A survey shows that 30% of college students own a car. If we randomly select 12 students, what is the probability that the number of car owners is between 2 and 5 inclusive?

$P(2 \leq X \leq 5) = \binom{12}{5}(0.3)^5(0.7)^7 - \binom{12}{2}(0.3)^2(0.7)^{10}$
$P(2 \leq X \leq 5) = \sum_{k=3}^{6} \binom{12}{k}(0.3)^k(0.7)^{12-k}$
$P(2 \leq X \leq 5) = \sum_{k=2}^{5} \binom{12}{k}(0.3)^k(0.7)^{12-k}$ (correct answer)
$P(2 \leq X \leq 5) = \binom{12}{3}(0.3)^3(0.7)^9$

Explanation: When you see a problem about randomly selecting individuals from a population with a known success rate, you're dealing with a binomial distribution. Here, each student either owns a car (probability 0.3) or doesn't (probability 0.7), and we're selecting 12 independent trials. For a binomial distribution, the probability of exactly k successes in n trials is

P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}

. To find the probability of getting between 2 and 5 car owners inclusive, you need to sum the individual probabilities for each possible value: k = 2, 3, 4, and 5. Answer C correctly shows this:

P(2 \leq X \leq 5) = \sum_{k=2}^{5} \binom{12}{k}(0.3)^k(0.7)^{12-k}

. This adds up

P(X=2) + P(X=3) + P(X=4) + P(X=5)

. Answer A attempts to use subtraction but incorrectly calculates

P(X=5) - P(X=2)

, which doesn't give you the probability of the range. Answer B shifts the summation limits to k = 3 through 6, missing k = 2 and incorrectly including k = 6. Answer D only calculates the probability for exactly 3 car owners, not the entire range from 2 to 5. Remember: for "between a and b inclusive" in binomial problems, you must sum all individual probabilities from k = a to k = b. Don't fall for shortcuts that subtract individual probabilities or shift the range boundaries.

Question 15

A multiple-choice test has 15 questions, each with 4 options. A student guesses randomly on all questions. What is the probability that the student gets at least 6 questions correct?

$\binom{15}{6}(0.25)^6(0.75)^9$
$\sum_{k=6}^{15} \binom{15}{k}(0.25)^k(0.75)^{15-k}$
$1 - \sum_{k=0}^{6} \binom{15}{k}(0.25)^k(0.75)^{15-k}$
$1 - \sum_{k=0}^{5} \binom{15}{k}(0.25)^k(0.75)^{15-k}$ (correct answer)

Explanation: When you encounter problems about repeated independent trials with fixed success probability, you're dealing with binomial probability. Here, each question is an independent trial with probability

p = 0.25

of success (guessing correctly) and

q = 0.75

of failure. The question asks for "at least 6 correct," meaning 6, 7, 8, ..., or 15 correct answers. You can approach this two ways: directly sum all favorable outcomes, or use the complement rule (often easier for "at least" problems). Using the complement rule: P(at least 6 correct) = 1 - P(fewer than 6 correct) = 1 - P(0, 1, 2, 3, 4, or 5 correct). This gives us

1 - \sum_{k=0}^{5} \binom{15}{k}(0.25)^k(0.75)^{15-k}

, which is answer choice D. Choice A only calculates the probability of getting exactly 6 questions correct, missing all other favorable outcomes (7, 8, 9, etc.). Choice B represents the direct approach—summing probabilities from 6 to 15 correct—which would also give the right answer, but this isn't among our options exactly as written. Choice C makes a critical error in the complement calculation. It subtracts P(0 through 6 correct) instead of P(0 through 5 correct), which would give us P(at least 7 correct) rather than P(at least 6 correct). For "at least k" problems in binomial distributions, always check whether using the complement (1 minus "fewer than k") is easier than direct calculation. Pay careful attention to whether the complement should exclude k itself or include it.

Question 16

In a binomial distribution with n = 20 and p = 0.25, which statement about the distribution's shape and characteristics is most accurate?

The distribution is approximately symmetric since np = 5 and n(1-p) = 15
The distribution is right-skewed since p < 0.5, with mode at k = 5 (correct answer)
The distribution is left-skewed since n(1-p) > np, with mode at k = 4
The distribution is approximately normal since both np > 5 and n(1-p) > 5

Explanation: With n=20 and p=0.25: np=5 and n(1-p)=15. Since p < 0.5, the distribution is right-skewed. The mode is approximately np = 5. Choice A is wrong because the distribution isn't symmetric when p ≠ 0.5. Choice C has the wrong skew direction. Choice D is wrong because np = 5 is borderline for normal approximation, and the skewness makes it inappropriate.

Question 17

In a binomial experiment with n = 100 and p = 0.04, what is the most appropriate method to calculate P(X ≥ 8)?

Use the normal approximation with continuity correction since np > 5 and n(1-p) > 5
Use the Poisson approximation since n is large and p is small with np = 4 (correct answer)
Calculate directly using the binomial formula since n = 100 is manageable
Use the normal approximation without continuity correction since the sample size is large

Explanation: With n=100 and p=0.04, we have np=4 and n(1-p)=96. Since n is large, p is small, and np is moderate, the Poisson approximation is most appropriate. Choice A is wrong because np=4 < 5, violating the normal approximation condition. Choice C is computationally intensive and unnecessary. Choice D is wrong for the same reason as A, plus it ignores the continuity correction.

Question 18

Consider a binomial random variable X with parameters n = 40 and p = 0.12. Using the normal approximation with continuity correction, what is the appropriate expression for P(X > 6)?

$P\left(Z > \frac{6 - 4.8}{\sqrt{4.224}}\right)$
$P\left(Z > \frac{6.5 - 4.8}{\sqrt{4.224}}\right)$ (correct answer)
$P\left(Z > \frac{7 - 4.8}{\sqrt{4.224}}\right)$
$P\left(Z > \frac{5.5 - 4.8}{\sqrt{4.224}}\right)$

Explanation: With n=40 and p=0.12: μ = np = 4.8, σ² = np(1-p) = 40(0.12)(0.88) = 4.224. For P(X > 6) with continuity correction, we need P(X ≥ 7), which becomes P(Z > (6.5-4.8)/√4.224). Choice A omits continuity correction. Choice C uses 7 instead of 6.5 for the correction. Choice D uses the wrong correction direction (5.5 instead of 6.5).

Question 19

In a binomial distribution with parameters $n = 20$ and $p = 0.4$ , what is the probability that the number of successes differs from the expected value by more than $2$ standard deviations?

$P(X < 3.09) + P(X > 12.91)$ where $X \sim \text{Binomial}(20, 0.4)$
$P(X \leq 3) + P(X \geq 13)$ where $X \sim \text{Binomial}(20, 0.4)$ (correct answer)
$P(X < 4) + P(X > 12)$ where $X \sim \text{Binomial}(20, 0.4)$
$P(X \leq 4) + P(X \geq 12)$ where $X \sim \text{Binomial}(20, 0.4)$

Explanation: Expected value μ = np = 20(0.4) = 8. Standard deviation σ = √(np(1-p)) = √(20(0.4)(0.6)) = √4.8 ≈ 2.19. Two standard deviations from mean: 8 ± 2(2.19) = 8 ± 4.38, giving interval (3.62, 12.38). Since X is discrete, we want P(X ≤ 3) + P(X ≥ 13). Choice A uses continuous boundaries. Choice C uses strict inequalities incorrectly. Choice D includes the boundary values that are within 2 standard deviations.