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Statistics and Distributions · Ages 16–17

Binomial Probability Practice Sheet

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For 11th-grade students. Verify the four binomial conditions, determine n, p, and k, and solve each problem step by step.

  1. A fair coin is flipped 4 times. What is n, and what is p for landing on heads?

  2. Calculate the binomial coefficient 6 choose 2.

  3. A fair 6-sided die is rolled 5 times. What is the probability of rolling no sixes at all?

  4. A multiple-choice quiz has 4 questions, each with 5 answer choices. If you guess randomly on all 4, what is the probability of getting exactly 2 correct?

  5. A machine produces parts that are defective 10% of the time. If a batch of 6 parts is inspected, what is the probability of finding exactly 1 defective part?

  6. Using the same machine from item 5 (defect rate 10%, sample of 6), calculate the probability of finding at least 1 defective part using the complement rule.

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Binomial Probability Practice Sheet

Answer key

  1. A fair coin is flipped 4 times. What is n, and what is p for landing on heads?

    n = 4 trials; p = 0.5 (or 1/2).

  2. Calculate the binomial coefficient 6 choose 2.

    6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15.

  3. A fair 6-sided die is rolled 5 times. What is the probability of rolling no sixes at all?

    (5 choose 0) * (1/6)^0 * (5/6)^5 = 1 * 1 * (3125 / 7776) = 3125/7776, or approximately 0.4019 (40.19%).

  4. A multiple-choice quiz has 4 questions, each with 5 answer choices. If you guess randomly on all 4, what is the probability of getting exactly 2 correct?

    (4 choose 2) * (0.2)^2 * (0.8)^2 = 6 * 0.04 * 0.64 = 0.1536 (15.36%).

  5. A machine produces parts that are defective 10% of the time. If a batch of 6 parts is inspected, what is the probability of finding exactly 1 defective part?

    (6 choose 1) * (0.1)^1 * (0.9)^5 = 6 * 0.1 * 0.59049 = 0.354294, or approximately 0.3543 (35.43%).

  6. Using the same machine from item 5 (defect rate 10%, sample of 6), calculate the probability of finding at least 1 defective part using the complement rule.

    P(at least 1) = 1 - P(0) = 1 - [(6 choose 0) * (0.1)^0 * (0.9)^6] = 1 - 0.531441 = 0.468559, or approximately 0.4686 (46.86%).