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Quantitative Aptitude · Theoretical Distributions

Binomial Distribution for CA Foundation

Updated 1 October 2026 · Fact-checked

The binomial distribution gives the probability of exactly r successes in n independent trials, each with the same success probability p. Use P(X = r) = nCr × p^r × q^(n−r), where q = 1 − p. Identify n, p and r, substitute them, and compute carefully.

Understand Binomial Distribution

Start with a Bernoulli trial. It is a single experiment with only two outcomes: success or failure. Tossing a coin once, checking one item as defective or not, or asking one customer whether they buy are all Bernoulli trials.

Now repeat the same trial n times. The binomial distribution tells you how likely it is to get exactly r successes in those n trials. The variable X = number of successes can take values 0, 1, 2, ..., n.

Why does the formula look the way it does? One particular sequence with r successes and (n − r) failures has probability p^r × q^(n−r), because the trials are independent. But the r successes can sit in any positions. The number of such arrangements is nCr. Multiply the two and you get the formula.

The distribution applies only when these conditions hold:

  • There is a fixed number of trials, n.
  • Each trial has only two outcomes: success or failure.
  • The trials are independent.
  • The probability of success p is the same in every trial.

If an exam question says items are drawn without replacement from a small lot, p changes from draw to draw. The binomial model does not fit exactly then. Read the wording for 'with replacement' or 'large population'.

Key formulas to remember

Binomial probability
P(X = r) = nCr × p^r × q^(n−r), r = 0, 1, 2, ..., n
n = number of trials, r = number of successes, p = probability of success, q = 1 − p.
Combination
nCr = n! ÷ [r! × (n − r)!]
Counts the ways to place r successes among n trials. nCr = nC(n−r).
Probability of failure
q = 1 − p
p + q = 1 in every trial.
Total probability
Σ P(X = r) for r = 0 to n equals 1
Use it to check answers or to find 'at least' by subtraction.
At least one success
P(X ≥ 1) = 1 − q^n
Faster than adding P(1) + P(2) + ... + P(n).
Mean and variance
Mean = np; Variance = npq
Standard deviation = √(npq). Since q < 1 for p > 0, variance npq < mean np. They are equal only in the trivial case p = 0.
Recurrence relation
P(r + 1) = [(n − r) ÷ (r + 1)] × (p ÷ q) × P(r)
Useful when you need several consecutive terms starting from P(0) = q^n.

How to solve Binomial Distribution questions

Use this method for any binomial question. It keeps your working short and your signs correct.

  1. 1Check the conditions: fixed n, two outcomes, independent trials, constant p.
  2. 2Define 'success' as the event the question counts, and write down n and p.
  3. 3Find q = 1 − p.
  4. 4Translate the wording into r: 'exactly 3' means r = 3, 'at least 2' means r ≥ 2, 'at most 2' means r ≤ 2.
  5. 5For a single r, compute nCr × p^r × q^(n−r).
  6. 6For ranges, add the needed terms. If the range is long, use the complement: 1 − P(the other values).
  7. 7Compare the result with the four options. Check that it lies between 0 and 1.

Quickest way: Complement and option-elimination method

When to use it: Use it in the 2-hour MCQ paper when a question asks for 'at least' or when the numbers look heavy.

  1. For 'at least one', compute 1 − q^n directly. Do not add terms.
  2. For 'at most' or 'at least' with small n, count the terms on each side and add the shorter side.
  3. Use symmetry for p = 0.5: P(r) = P(n − r), and the coefficients nCr are the row of Pascal's triangle (for n = 5: 1, 5, 10, 10, 5, 1).
  4. Keep fractions as powers of 2 or 10 and avoid decimals until the end.
  5. Eliminate options: the answer must be between 0 and 1, and P(at least r) must be at least as large as P(exactly r).
  6. If n is large and p is awkward, with several terms to add, skip and return later. A wrong answer costs 0.25 marks.

Common mistakes in Binomial Distribution

  • Forgetting the nCr term and writing only p^r × q^(n−r).

    Students picture one fixed order of successes and failures.

    Fix: Always ask 'in how many positions can the successes occur?' and multiply by nCr.

  • Using p for the wrong event, such as taking p as the defective rate when the question counts good items.

    The word 'success' sounds positive, so students assume it means the favourable thing.

    Fix: Success is whatever the question counts. Write 'success = ...' in your rough work before choosing p.

  • Misreading 'at least' and 'at most'.

    Both phrases involve ranges, and students rush the inequality.

    Fix: Write the range as r ≥ or r ≤ with actual numbers before calculating. 'At least 2' includes 2.

  • Applying the formula when trials are not independent or p changes.

    Students match keywords instead of checking conditions.

    Fix: Check the four conditions first. Draws without replacement from a small lot are not binomial.

  • Errors in powers, such as (0.9)^2 written as 0.18 or q^(n−r) written as q^r.

    Mental arithmetic under time pressure.

    Fix: Check that the two powers add up to n. Square decimals separately: 0.9² = 0.81.

  • Adding up probabilities and getting a value above 1.

    A wrong p or q was used, or the same term was counted twice.

    Fix: Use the check that any probability lies between 0 and 1, and that p + q = 1.

Worked examples

Example 1

A fair coin is tossed 5 times. What is the probability of getting exactly 3 heads? (A) 5/16 (B) 3/8 (C) 1/4 (D) 5/32

Show the solution
  1. Trials are independent with two outcomes, so the binomial applies. n = 5, p = 1/2, q = 1/2, r = 3.
  2. P(X = 3) = 5C3 × (1/2)^3 × (1/2)^2.
  3. 5C3 = 10 and (1/2)^5 = 1/32.
  4. P = 10 × 1/32 = 10/32 = 5/16.

Answer: (A) 5/16

Example 2

The probability that a machine produces a good item is 0.2 in each independent trial. In 4 trials, what is the probability of at least one good item? (A) 0.4096 (B) 0.5904 (C) 0.8192 (D) 0.1536

Show the solution
  1. Success = good item. n = 4, p = 0.2, q = 0.8.
  2. 'At least one' is the complement of 'none'.
  3. P(X = 0) = q^4 = 0.8^4. First 0.8² = 0.64, then 0.64² = 0.4096.
  4. P(X ≥ 1) = 1 − 0.4096 = 0.5904.

Answer: (B) 0.5904

Example 3

Each item produced by a factory is defective with probability 0.1, independently. In a sample of 3 items, what is the probability that exactly 2 are defective? (A) 0.243 (B) 0.081 (C) 0.027 (D) 0.010

Show the solution
  1. Success = defective. n = 3, p = 0.1, q = 0.9, r = 2.
  2. P(X = 2) = 3C2 × (0.1)^2 × (0.9)^1.
  3. 3C2 = 3, (0.1)^2 = 0.01, (0.9)^1 = 0.9.
  4. P = 3 × 0.01 × 0.9 = 0.027.

Answer: (C) 0.027

Exam tips

  • Questions are usually direct: given n and p, find P(exactly r) or an 'at least' probability. Practise these until the setup takes seconds.
  • Check for the phrase 'independent' or 'with replacement'. It confirms the binomial model is meant.
  • Expect the same data to be used in the mean-and-variance questions, so remember np and npq with this topic.
  • Keep a table of nCr for n up to 10 in your head. It saves most of the calculation time.
  • Option traps are common: the answer with the nCr missing, or with p and q swapped, will often appear among the choices. Recheck which one you computed.

Practice questions from Theoretical Distributions

Binomial Distribution: frequently asked questions

What are the conditions of the binomial distribution?

There must be a fixed number of trials n. Each trial has only two outcomes, success or failure. The trials are independent, and the probability of success p stays the same in every trial.

What is the binomial distribution formula?

P(X = r) = nCr × p^r × q^(n−r), where q = 1 − p and r runs from 0 to n. It gives the probability of exactly r successes in n trials.

How do I find 'at least' probabilities quickly?

Use the complement. For at least one success, the answer is 1 − q^n. For other ranges, add whichever side has fewer terms and subtract from 1 if needed.

What is a Bernoulli trial?

It is a single experiment with two outcomes, success with probability p and failure with probability q = 1 − p. A binomial distribution describes n independent Bernoulli trials with the same p.

Is the variance of a binomial distribution larger than its mean?

No. The mean is np and the variance is npq. Since q is less than 1 (when p is not 0), the variance is smaller than the mean.