Statistics: The Binomial Distribution
Modeling repeated trials with two possible outcomes
Statistics: The Binomial Distribution
Modeling repeated trials with two possible outcomes
Statistics - Grade 9-12
- 1
A random variable X counts the number of successes in 10 independent trials. Each trial has exactly two outcomes, and the probability of success is 0.30 for every trial. Explain why X can be modeled with a binomial distribution and identify n and p.
Check the four binomial conditions: fixed number, independent trials, two outcomes, and constant probability.
X can be modeled with a binomial distribution because there is a fixed number of trials, the trials are independent, each trial has two outcomes, and the probability of success stays the same. Here, n = 10 and p = 0.30. - 2
A basketball player makes 70% of her free throws. If she shoots 6 free throws, what is the probability that she makes exactly 4 of them?
Let X be the number of made free throws. X follows a binomial distribution with n = 6 and p = 0.70. The probability is P(X = 4) = C(6,4)(0.70)^4(0.30)^2 = 0.3241. - 3
A multiple-choice quiz has 8 questions, each with 4 answer choices. A student guesses on every question. What is the probability that the student gets exactly 3 questions correct?
A correct guess is the success, so p is 1 out of 4.
The probability of a correct guess is p = 0.25, and n = 8. The probability is P(X = 3) = C(8,3)(0.25)^3(0.75)^5 = 0.2076. - 4
A factory knows that 5% of its light bulbs are defective. In a random sample of 20 bulbs, what is the probability that exactly 2 are defective?
Let X be the number of defective bulbs. X follows a binomial distribution with n = 20 and p = 0.05. The probability is P(X = 2) = C(20,2)(0.05)^2(0.95)^18 = 0.1887. - 5
A coin is flipped 12 times. What is the probability of getting at least 10 heads?
At least 10 means 10, 11, or 12 successes.
Let X be the number of heads. X follows a binomial distribution with n = 12 and p = 0.50. The probability is P(X at least 10) = P(X = 10) + P(X = 11) + P(X = 12) = 0.0193. - 6
A survey finds that 60% of students at a school prefer digital textbooks. If 15 students are randomly selected, what is the probability that fewer than 5 prefer digital textbooks?
Let X be the number of students who prefer digital textbooks. X follows a binomial distribution with n = 15 and p = 0.60. The probability is P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0338. - 7
A game has a 20% chance of winning on each play. If a person plays 9 times, what is the probability that the person wins at least once?
Use the complement of winning at least once, which is winning zero times.
Let X be the number of wins. X follows a binomial distribution with n = 9 and p = 0.20. It is easier to use the complement: P(X at least 1) = 1 - P(X = 0) = 1 - (0.80)^9 = 0.8658. - 8
A baseball player has a 0.280 probability of getting a hit in each at-bat. In 5 at-bats, what is the probability that he gets no hits?
Let X be the number of hits. X follows a binomial distribution with n = 5 and p = 0.280. The probability is P(X = 0) = C(5,0)(0.280)^0(0.720)^5 = 0.1935. - 9
A binomial random variable X has n = 40 and p = 0.25. Find the mean and standard deviation of X.
For a binomial distribution, the mean is np and the standard deviation is sqrt(np(1 - p)).
The mean is mu = np = 40(0.25) = 10. The standard deviation is sigma = sqrt(np(1 - p)) = sqrt(40(0.25)(0.75)) = sqrt(7.5), which is about 2.7386. - 10
A website reports that 12% of visitors click on a certain ad. If 50 visitors come to the site, what is the expected number of visitors who click the ad?
Let X be the number of visitors who click the ad. Since X is binomial with n = 50 and p = 0.12, the expected value is E(X) = np = 50(0.12) = 6. The expected number of visitors who click the ad is 6. - 11
A quality control inspector samples 30 items from a production line where each item has a 3% chance of being flawed. What is the probability that the sample contains 1 or fewer flawed items?
Add the probability of 0 flawed items and the probability of 1 flawed item.
Let X be the number of flawed items. X follows a binomial distribution with n = 30 and p = 0.03. The probability is P(X <= 1) = P(X = 0) + P(X = 1) = (0.97)^30 + C(30,1)(0.03)(0.97)^29 = 0.7723. - 12
A teacher says the number of students who pass a test in a class of 25 can always be modeled by a binomial distribution if the overall pass rate is 80%. Explain one reason this claim might not be valid.
Think about whether every student has the same probability of passing and whether students' results are independent.
The claim might not be valid because the trials may not be independent or the probability of passing may not be the same for every student. For example, students may have different preparation levels, so the probability of success may vary.