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Binomial Distribution formulas

13 formulas and 12 common traps for NDA Mathematics Binomial Distribution, grouped by subtopic.

Full notes with worked examples

The Binomial Setting and Computing Probabilities

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Bernoulli Trials: One Success, One Failure

Failure complements success

q=1−p,p+q=1q = 1 - p, \qquad p + q = 1

Reading p and the Success Event from the Story

Odds to probability

odds a:b ⟹ p=aa+b\text{odds } a : b \ \Longrightarrow\ p = \dfrac{a}{a + b}

The Binomial Probability Formula

Probability of exactly k successes

P(X=k)=(nk)pkq n−kP(X = k) = \binom{n}{k} p^{k} q^{\,n-k}
  • nnnumber of trials
  • kknumber of successes counted
  • ppsuccess probability per trial
  • qqfailure probability, 1 − p

At Least One via the Complement

At least one success

P(X≥1)=1−qnP(X \ge 1) = 1 - q^{n}

Cumulative Probabilities: Summing the Tail

At least two successes

P(X≥2)=1−P(X=0)−P(X=1)P(X \ge 2) = 1 - P(X = 0) - P(X = 1)

The Complementary Count Y = n − X

Swapping successes for failures

X∼B(n,p) ⟹ n−X∼B(n, 1−p)X \sim B(n, p) \ \Longrightarrow\ n - X \sim B(n,\, 1 - p)

Common traps

'Until the first success' is not binomial

If the number of trials is not fixed in advance — e.g. 'toss until a head appears' — then nn is random and the binomial formula does not apply. Binomial needs a pre-set nn.

'Thrice as likely' is 3 : 1, not p = 3

Odds split the whole into parts. 'Heads thrice as likely as tails' means 3:13 : 1 out of 4 parts, so p=34p = \tfrac{3}{4} — never read it as p=3p = 3 or p=13p = \tfrac{1}{3}.

Match the exponents to the success/failure counts

The power on pp is the number of successes kk; the power on qq is n−kn - k. Swapping them — e.g. p n−kqkp^{\,n-k} q^{k} — is the single most common slip, especially when p≠qp \ne q.

'At most' can also flip to a complement

'At most 4 tails in 5 tosses' has five terms the long way, but its complement 'exactly 5 tails' is one term: 1−(12)5=31321 - (\tfrac12)^5 = \tfrac{31}{32}. Read the count and complement whichever side is shorter.

Count the terms before you sum

For 'at least kk', the direct sum runs j=kj = k to nn and the complement runs j=0j = 0 to k−1k-1. Pick the shorter list — and never forget a boundary term (the j=nj = n all-success term is easy to drop).

n stays the same — only p flips

A frequent distractor halves nn or keeps pp unchanged. The complementary count n−Xn - X keeps the SAME nn and only swaps p↔qp \leftrightarrow q.

Mean, Variance, and Recovering the Parameters

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Why the Mean Is np and the Variance npq

Mean and variance of one trial

E(I)=p,Var⁡(I)=pqE(I) = p, \qquad \operatorname{Var}(I) = pq

Mean, Variance, and Standard Deviation

The three summary measures

μ=np,σ2=npq,σ=npq\mu = np, \qquad \sigma^2 = npq, \qquad \sigma = \sqrt{npq}

Recovering n and p from the Moments

Divide variance by mean to get q

q=σ2μ=npqnpq = \dfrac{\sigma^2}{\mu} = \dfrac{npq}{np}

When You Are Given a Relation, Not the Values

Mean equals c times variance

np=c (npq) ⟹ q=1cnp = c\,(npq) \ \Longrightarrow\ q = \dfrac{1}{c}

Finding p from a Probability Equation

Ratio of two probabilities

P(X=b)P(X=a)=(nb)(na) p b−a q a−b\dfrac{P(X=b)}{P(X=a)} = \dfrac{\binom{n}{b}}{\binom{n}{a}}\, p^{\,b-a}\, q^{\,a-b}

Variance Is Unchanged by Y = n − X

Variance survives the swap

Var⁡(n−X)=npq=Var⁡(X)\operatorname{Var}(n - X) = npq = \operatorname{Var}(X)

The Symmetric Case: Mean = n/2 when p = ½

Symmetric binomial mean

p=12 ⟹ μ=n2p = \tfrac12 \ \Longrightarrow\ \mu = \dfrac{n}{2}

Common traps

Standard deviation is √(npq), not npq

Questions love to give the standard deviation and call it the variance, or vice versa. Square the SD to get the variance before using σ2=npq\sigma^2 = npq — e.g. SD =2= \sqrt2 means variance =2= 2.

Variance over mean gives q, not p

The ratio σ2/μ=q\sigma^2/\mu = q (the FAILURE probability). Forgetting the final p=1−qp = 1 - q step lands you on the complement and the wrong nn.

Cancel np, do not cancel the wrong factor

From np=c npqnp = c\,npq, the surviving factor is qq (giving q=1/cq = 1/c). Cancelling to leave pp instead — a common slip — inverts the answer.

Use coefficient symmetry before brute force

Spotting (64)=(62)\binom{6}{4} = \binom{6}{2} cancels the coefficients in one step. Expanding them numerically still works but invites arithmetic slips — and forgetting to take the positive root of 9p2=(1−p)29p^2 = (1-p)^2 loses the intended answer.

Variance does not flip; the mean does

Y=n−XY = n - X leaves the variance equal to Var⁡(X)\operatorname{Var}(X), but its mean becomes n−npn - np. Do not 'adjust' the variance for the swap — only the mean changes.

Symmetry needs p = ½, not just 'two outcomes'

The mean is n/2n/2 ONLY when p=q=12p = q = \tfrac12. With a biased trial the peak shifts to npnp — do not default to n/2n/2 unless success and failure are equally likely.

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