EAPCET Mathematics · Intermediate second year · Paper IIA

Random Variables and Probability Distributions

This chapter treats outcomes as random variables, computes their mean and variance, and introduces the binomial and Poisson distributions.

Subject
Mathematics
Taught in
Intermediate second year · Paper IIA
Mathematics in the paper
80 of 160 questions

Key formulas and results

  • ∑P(xi)=1\sum P(x_{i}) = 1
  • Mean μ=∑xiP(xi)\mu = \sum x_{i} P(x_{i}); variance =∑xi2P(xi)−μ2= \sum x_{i}^{2} P(x_{i}) - \mu^{2}
  • Binomial: P(X=r)=nCr prqn−rP(X = r) = {}^{n}C_{r}\, p^{r} q^{n-r}, with mean npnp and variance npqnpq
  • Poisson: P(X=r)=e−λλrr!P(X = r) = \dfrac{e^{-\lambda}\lambda^{r}}{r!}, with mean and variance both λ\lambda
  • Binomial: P(X≥r)=∑k=rnnCk pkqn−kP(X \ge r) = \displaystyle\sum_{k=r}^{n} {}^{n}C_{k}\, p^{k} q^{n-k} and P(X≤r)=∑k=0rnCk pkqn−kP(X \le r) = \displaystyle\sum_{k=0}^{r} {}^{n}C_{k}\, p^{k} q^{n-k}
  • Standard deviation of a binomial variate =npq= \sqrt{npq}

What this chapter covers

  • Random variables and probability distributions
  • Mean and variance
  • Binomial distribution
  • Poisson distribution

How it is usually asked

  • Find an unknown constant in a distribution
  • Compute binomial probabilities
  • Find n and p from a given mean and variance
  • Poisson probabilities

Understanding Random Variables and Probability Distributions

A random variable assigns a number to each outcome of an experiment, such as the number of heads in three tosses. Its probability distribution lists each possible value with its probability, and these probabilities must be non-negative and add to 1. That condition is how questions with an unknown constant k in the table are solved.

The mean, or expected value, is μ = Σxᵢ P(xᵢ), the long-run average value. The variance is Σxᵢ² P(xᵢ) − μ², which measures spread around the mean in the same way as for data. These two formulas apply to any discrete probability distribution and are worth practising on small tables until they become automatic.

Two distributions are named in the syllabus. The binomial distribution counts successes in n independent trials, each with success probability p: P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ, with mean np and variance npq, so its variance is always less than its mean. The Poisson distribution, P(X = r) = e^(−λ) λʳ/r!, models rare events and has mean and variance both equal to λ.

Worked example

EAPCET-style question

A binomial distribution has mean 4 and variance 3. Its parameters are

  1. (1)n=16, p=14n = 16,\ p = \tfrac{1}{4}
  2. (2)n=16, p=34n = 16,\ p = \tfrac{3}{4}
  3. (3)n=12, p=13n = 12,\ p = \tfrac{1}{3}
  4. (4)n=8, p=12n = 8,\ p = \tfrac{1}{2}

Solution

  1. np=4np = 4 and npq=3npq = 3, so q=34q = \tfrac{3}{4}.
  2. p=1−q=14p = 1 - q = \tfrac{1}{4}.
  3. n=4p=16n = \dfrac{4}{p} = 16.
  4. Check: np=4np = 4 and npq=16×14×34=3npq = 16 \times \tfrac{1}{4} \times \tfrac{3}{4} = 3.

Answer: n=16, p=14n = 16,\ p = \tfrac{1}{4}

Shortcuts and results to remember

  • For a binomial variate, variance npqnpq is always less than the mean npnp, since q<1q < 1; a question giving variance > mean has no binomial solution.

More practice questions

Practice question 1

For a binomial variate, mean−variance=1\text{mean} - \text{variance} = 1 and (mean)2−(variance)2=11(\text{mean})^{2} - (\text{variance})^{2} = 11. The number of trials nn is

  1. (1)36
  2. (2)6
  3. (3)30
  4. (4)11

Solution

  1. np−npq=1np - npq = 1 and (np)2−(npq)2=11(np)^{2} - (npq)^{2} = 11
  2. Divide: np+npq=11np + npq = 11
  3. Adding: 2np=122np = 12, so np=6np = 6 and npq=5npq = 5
  4. q=56q = \tfrac{5}{6}, p=16p = \tfrac{1}{6}, so n=61/6=36n = \dfrac{6}{1/6} = 36

Answer: 36

Practice question 2

A pair of dice is thrown 5 times and getting a doublet is a success. The mean and variance of the number of successes are

  1. (1)56\tfrac{5}{6} and 2536\tfrac{25}{36}
  2. (2)256\tfrac{25}{6} and 2536\tfrac{25}{36}
  3. (3)56\tfrac{5}{6} and 536\tfrac{5}{36}
  4. (4)16\tfrac{1}{6} and 536\tfrac{5}{36}

Solution

  1. Doublets: 6 of the 36 outcomes, so p=16p = \tfrac{1}{6} and q=56q = \tfrac{5}{6}
  2. Mean =np=5×16=56= np = 5 \times \tfrac{1}{6} = \tfrac{5}{6}
  3. Variance =npq=5×16×56=2536= npq = 5 \times \tfrac{1}{6} \times \tfrac{5}{6} = \tfrac{25}{36}

Answer: 56\tfrac{5}{6} and 2536\tfrac{25}{36}

Common mistakes to avoid

  • Accepting a binomial distribution whose variance exceeds its mean; since q < 1, npq < np always.
  • Using Σxᵢ² P(xᵢ) as the variance without subtracting μ².
  • Writing the Poisson probability with e^λ instead of e^(−λ).
  • Computing P(X ≥ 1) by adding many terms instead of 1 − P(X = 0).

How to prepare it

If you know a binomial mean and variance, q = variance/mean gives p and n in two steps.

Once the formulas feel familiar, test them the way the exam does: practise Random Variables and Probability Distributions questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Mathematics chapters

See every chapter in the EAPCET Mathematics syllabus.

Practise Random Variables and Probability Distributions chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Random Variables and Probability Distributions part of the EAPCET Mathematics syllabus?
Yes. Random Variables and Probability Distributions is taught in Intermediate second year · Paper IIA Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Random Variables and Probability Distributions for EAPCET?
If you know a binomial mean and variance, q = variance/mean gives p and n in two steps.
How many Mathematics questions are there in EAPCET?
80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.
What mistakes do students make in Random Variables and Probability Distributions?
Accepting a binomial distribution whose variance exceeds its mean; since q < 1, npq < np always. Using Σxᵢ² P(xᵢ) as the variance without subtracting μ². Writing the Poisson probability with e^λ instead of e^(−λ). Computing P(X ≥ 1) by adding many terms instead of 1 − P(X = 0).