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JEE Mains Maths · Probability

Binomial Distribution

The number of successes in n independent trials with the same success probability: P(X = r) = C(n, r) p^r q^(n − r), with mean np and variance npq.

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Why this matters

Twenty PYQs, and 2022 alone has nine. Thirteen compute a probability — exactly r, at least r, or a ratio of two terms — and seven recover n and p from the mean and variance first. Two ideas cover the page.

Concept 1 of 2: Binomial probabilities

For nn independent trials with success probability pp, exactly rr successes happen with probability (nr)prqn−r\binom nrp^rq^{n-r}. 'At least rr' adds the top terms or subtracts the bottom ones, whichever is shorter. A ratio of two terms cancels the binomial coefficients when they are symmetric, leaving a power of qp\frac qp; an equation like P(X=3)=5P(X=4)P(X=3)=5P(X=4) gives qp\frac qp directly.

Definition

  • P(X=r)=(nr)prqn−rP(X=r)=\binom nrp^rq^{n-r}, q=1−pq=1-p.
  • P(X=r+1)P(X=r)=n−rr+1⋅pq\frac{P(X=r+1)}{P(X=r)}=\frac{n-r}{r+1}\cdot\frac pq.
  • P(X=r)=P(X=s)P(X=r)=P(X=s) with p=12p=\frac12 means (nr)=(ns)\binom nr=\binom ns, so n=r+sn=r+s.
  • At least one: 1−qn1-q^n.

Binomial probability

P(X=r)=(nr) prq n−rP(X=r)=\binom nr\,p^{r}q^{\,n-r}

Worked example

A fair coin is tossed 6 times. Find the probability of at least 5 heads.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 6 April 2023 · Q62Moderate

Example 1 · Probability · Binomial Distribution

A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If the probability of at least 4 successes is k311\frac{k}{3^{11}}, then kk is equal to

Identify the trial and its p

When a pair of dice is thrown, one throw of the pair is one trial; its success probability comes from counting outcomes of the pair (sum 5 has probability 19\frac19), not from one die.
Drill 12 more on binomial probabilities

Concept 2 of 2: Recovering n and p from the mean and variance

Mean npnp and variance npqnpq give qq as their ratio: q=variancemeanq=\frac{\text{variance}}{\text{mean}}. If their sum and product are given instead, they are the roots of a quadratic; the larger root is the mean, since q<1q<1. Then p=1−qp=1-q and n=meanpn=\frac{\text{mean}}p.

Definition

  • Mean npnp, variance npqnpq, so q=npqnpq=\frac{npq}{np}.
  • Mean −- variance =np2=np^2.
  • Sum SS and product PP given: mean and variance are the roots of t2−St+P=0t^2-St+P=0, mean the larger.
  • Then compute the requested probability with the recovered n,pn,p.

Moments

E(X)=np,Var⁡(X)=npqE(X)=np,\qquad\operatorname{Var}(X)=npq

Worked example

A binomial variable has mean 6 and variance 4. Find nn and pp.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 26 July 2022 · Q168Moderate

Example 2 · Probability · Binomial Distribution

Let XX be a binomially distributed random variable with mean 4 and variance 43\frac{4}{3}. Then 54P(X≤2)54P(X \leq 2) is equal to

The mean is the larger root

Since q<1q<1, the variance is less than the mean. When the two come from a quadratic, assigning the smaller root to the mean gives q>1q>1, which is impossible.
Drill 6 more on recovering n and p from the mean and variance

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

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