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Binomial Distribution Reference cheat sheet - grade 10-12

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Statistics Grade 10-12

Binomial Distribution Reference Cheat Sheet

A printable reference covering binomial conditions, probability formulas, mean, variance, standard deviation, cumulative probability, and normal approximation for grades 10-12.

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The binomial distribution models the number of successes in a fixed number of repeated trials. Students need this reference because binomial problems appear often in probability, statistics, biology, business, and test preparation. This cheat sheet helps you identify when a situation is binomial and choose the correct formula quickly.

The most important ideas are the four binomial conditions, the probability formula, and the meaning of the parameters nn and pp. A binomial random variable is written X∼B(n,p)X \sim B(n,p), where nn is the number of trials and pp is the probability of success on each trial. Key summary formulas include the mean μ=np\mu = np, variance σ2=np(1−p)\sigma^2 = np(1-p), and standard deviation σ=np(1−p)\sigma = \sqrt{np(1-p)}.

Key Facts

  • A binomial setting has a fixed number of trials nn, only two outcomes per trial, a constant success probability pp, and independent trials.
  • If XX counts the number of successes in nn trials with success probability pp, then X∼B(n,p)X \sim B(n,p).
  • The binomial probability formula is P(X=k)=(nk)pk(1−p)n−kP(X=k)=\binom{n}{k}p^k(1-p)^{n-k} for k=0,1,2,…,nk=0,1,2,\ldots,n.
  • The combination formula is (nk)=n!k!(n−k)!\binom{n}{k}=\frac{n!}{k!(n-k)!}, which counts the number of ways to choose kk successes from nn trials.
  • The mean of a binomial distribution is μ=np\mu = np.
  • The variance of a binomial distribution is σ2=np(1−p)\sigma^2 = np(1-p).
  • The standard deviation of a binomial distribution is σ=np(1−p)\sigma = \sqrt{np(1-p)}.
  • A normal approximation is often reasonable when np≥10np \ge 10 and n(1−p)≥10n(1-p) \ge 10.

Vocabulary

Binomial distribution
A probability distribution for the number of successes in a fixed number of independent trials with the same success probability.
Trial
One repetition of a random experiment, such as flipping a coin once or asking one person a survey question.
Success
The outcome being counted in a binomial problem, even if it is not a positive or desirable result.
Parameter
A number that defines a probability model, such as nn for the number of trials and pp for the probability of success.
Combination
A count of selections where order does not matter, written as (nk)\binom{n}{k}.
Cumulative probability
The probability of getting a value within a range, such as P(X≤k)P(X \le k) or P(X≥k)P(X \ge k).

Common Mistakes to Avoid

  • Using the binomial formula when trials are not independent is wrong because the probability of success must stay the same from trial to trial.
  • Confusing pp and 1−p1-p changes the meaning of success and failure, so always define the success outcome before substituting into P(X=k)=(nk)pk(1−p)n−kP(X=k)=\binom{n}{k}p^k(1-p)^{n-k}.
  • Forgetting the combination factor (nk)\binom{n}{k} gives the probability of one specific order only, not all possible orders with kk successes.
  • Using P(X=k)P(X=k) when the question asks for at least or at most is wrong because cumulative wording requires adding several probabilities, such as P(X≤k)P(X \le k) or P(X≥k)P(X \ge k).
  • Applying the normal approximation when np<10np < 10 or n(1−p)<10n(1-p) < 10 can be inaccurate because the binomial distribution may be too skewed.

Practice Questions

  1. 1 A fair coin is flipped 88 times. If XX is the number of heads, find P(X=5)P(X=5).
  2. 2 A basketball player makes 70%70\% of free throws. If the player takes 1010 free throws, find the mean and standard deviation of the number made.
  3. 3 A multiple-choice quiz has 1212 questions with 44 choices each. If a student guesses on every question, what is P(X≥10)P(X \ge 10), where XX is the number correct?
  4. 4 A factory samples items without replacement from a small batch. Explain why this situation may not satisfy the binomial conditions.

Understanding Binomial Distribution Reference

A binomial model is built by turning each trial into an indicator. An indicator has value one for a success and zero for a failure. The total count is found by adding these indicator values across all trials.

This explains why the average count is number of trials times success probability. Each trial contributes its probability of success to the long-run total. It also explains why the spread depends on both the chance of success and the chance of failure.

Counts vary most when success is near one half. When success is very rare or very common, most results cluster closer to zero or to the total number of trials.

The combination part of a binomial probability is easy to misuse. First find the chance of one particular arrangement. For example, with five trials and two successes, one arrangement might be success, failure, success, failure, failure.

Its probability comes from multiplying two success probabilities and three failure probabilities. However, the two successes could occur in many different positions. Every arrangement with two successes has the same probability when the trials have the same success chance.

The combination counts those arrangements. It does not represent another probability. Students should calculate the number of success positions carefully before multiplying.

Many tasks ask for a range rather than one exact count. Phrases such as at most, fewer than, at least, and more than tell you which probabilities to combine. At most three includes zero through three.

At least three includes three through the largest possible count. A complement can make the work shorter. For instance, the probability of at least one success equals one minus the probability of zero successes.

This method is especially useful for rare events. A calculator can find cumulative probabilities, but students should still write the event in words first. This prevents common errors involving included endpoints.

Real data do not always fit the model perfectly. Repeated draws from a deck without replacement are not independent, because each draw changes what remains. Survey responses can be linked when friends influence one another.

A factory may have changing conditions during a shift, so the defect probability may not stay constant. In these cases, a binomial answer can be misleading even if the arithmetic is correct. The normal approximation is useful when a count distribution has enough room on both sides of its center.

It replaces separate bars for whole-number counts with a smooth curve. Since a normal curve is continuous, a continuity correction improves the estimate.

For a count of at most k, use the boundary halfway between k and the next whole number. Students should check the two size conditions before using this shortcut, especially when the success probability is close to zero or one.