Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
How to Use This Page
This page is your complete study companion for Chapter 11 — Probability Distributions. Start with the Key Concepts table to refresh definitions, then work through the MCQs, fill-in-the-blanks, and true-or-false sections for quick practice. Use the match-the-following exercise to strengthen associations. The brief and detailed answer sections mirror the 2-mark and 5-mark questions you will face in the board exam. Finally, review the Common Mistakes and Exam Checklist before every test to avoid losing marks.
Chapter 11 — Exercise Solutions Guide
Probability distribution questions become easy when you organise values in a table.
Key Concepts (Quick Recap)
| Term | Definition |
|---|---|
| Random variable | A variable whose value is determined by the outcome of a random experiment |
| Discrete random variable | A random variable that takes countable values (0, 1, 2, ...) |
| Continuous random variable | A random variable that takes any value in an interval |
| Probability mass function (PMF) | P(X = x) for each value x of a discrete random variable; all values are non-negative and sum to 1 |
| Cumulative distribution function | F(x) = P(X <= x); the probability that X takes a value less than or equal to x |
| Expected value (Mean) | E(X) = sum of x times P(X = x); the average value of the random variable |
| Variance | Var(X) = E(X²) - [E(X)]²; measures the spread of the distribution |
| Standard deviation | SD(X) = sqrt(Var(X)); square root of the variance |
| Binomial distribution | A discrete distribution with parameters n (trials) and p (probability of success) |
| Poisson distribution | A discrete distribution for rare events with parameter lambda (average rate) |
| Normal distribution | A continuous distribution with parameters mu (mean) and sigma (SD); bell-shaped curve |
| Probability condition | All P(x) >= 0 and sum of all P(x) = 1 |
Formula Recap
| Concept | Formula |
|---|---|
| Probability condition | Sum of p(x) = 1 |
| Mean | E(X) = sum xp(x) |
| Variance | Var(X) = E(X²) - [E(X)]² |
| Binomial probability | P(X = x) = nCx p^x q^(n-x) |
| Binomial mean | np |
| Binomial variance | npq |
| Poisson probability | P(X = x) = (e^(-lambda) lambda^x) / x! |
| Poisson mean and variance | Both equal lambda |
I. Choose the Best Answer (MCQs)
Corrected Answer: (b) 80/243
Corrected Answer: (a) 5/16 (P(X=3) = C(5,3)(1/2)^5 = 10/32 = 5/16.)
II. Fill in the Blanks
III. True or False
1. The sum of all probabilities in a discrete probability distribution must equal 1. True. (This is a fundamental axiom of probability distributions.)
2. The variance of a random variable can be negative. False. Correction: Variance is always non-negative because it is the expected value of a squared quantity (X - mu)².
3. In a binomial distribution, the probability of failure is q = 1 + p. False. Correction: The probability of failure is q = 1 - p, where p is the probability of success.
4. The mean and variance of a Poisson distribution are always equal. True. (Both the mean and variance of a Poisson distribution equal the parameter lambda.)
5. The standard deviation is always less than or equal to the variance. False. Correction: The standard deviation is the square root of the variance. For variance < 1, SD > variance; for variance > 1, SD < variance.
6. A binomial distribution is symmetric when p = 1/2. True. (When p = q = 1/2, the binomial distribution is symmetric about the mean n/2.)
7. E(X + Y) = E(X) + E(Y) for any two random variables X and Y. True. (The expectation operator is always linear, regardless of independence.)
8. The probability of each outcome in a discrete probability distribution can be greater than 1. False. Correction: Each probability P(X = x) must satisfy 0 <= P(X = x) <= 1. No individual probability can exceed 1.
IV. Match the Following
| Column A | Column B |
|---|---|
| 1. Mean of binomial | (c) np |
| 2. Variance of binomial | (e) npq |
| 3. Mean of Poisson | (a) lambda |
| 4. Variance of Poisson | (a) lambda |
| 5. E(X²) - [E(X)]² | (f) Variance |
| 6. Sum of all P(x) | (b) 1 |
| 7. Standard deviation | (g) sqrt(Var(X)) |
| 8. q in binomial | (d) 1 - p |
Probability Distribution Pattern
1. List all values of X. 2. Write corresponding probabilities. 3. Check that total probability is 1. 4. Calculate xp(x) for mean. 5. Calculate x²p(x) for variance.
Binomial Distribution Pattern
Use binomial distribution when:
- There are fixed number of trials.
- Each trial has two outcomes.
- Probability of success is constant.
- Trials are independent.
Then:
Worked Examples
Example 1 — Mean of a discrete distribution
X takes values 0, 1, 2 with probabilities 0.2, 0.5, 0.3. Find E(X).
Example 2 — Variance
For the same distribution, find Var(X).
Example 3 — Binomial distribution
A fair coin is tossed 10 times. Find the mean and variance of the number of heads.
Example 4 — Binomial probability
Find P(X = 2) for n = 5 trials, p = 1/3.
Example 5 — Find the value of k
If the probability distribution of X is given by P(X = 0) = k, P(X = 1) = 2k, P(X = 2) = 3k, P(X = 3) = 2k, find k.
V. Answer Briefly (2-Mark Questions)
1. What is a random variable? Distinguish between discrete and continuous random variables.
A random variable is a variable that takes different values based on the outcome of a random experiment. A discrete random variable takes countable values (like 0, 1, 2, 3), and its probabilities are given by a probability mass function P(X = x). A continuous random variable takes any value in an interval (like heights or weights), and its probabilities are given by a probability density function f(x) where P(a <= X <= b) = integral from a to b of f(x) dx.
2. State the conditions for a binomial distribution.
A binomial distribution B(n, p) applies when: (i) there are a fixed number n of independent trials, (ii) each trial has exactly two outcomes (success or failure), (iii) the probability of success p is constant for each trial, and (iv) the trials are independent of each other. The probability of exactly r successes is P(X = r) = nCr p^r q^(n-r) where q = 1 - p.
3. Define expected value and variance of a random variable.
The expected value (mean) of a discrete random variable X is E(X) = sum of x times P(X = x) over all values of X. It represents the average outcome over many repetitions. The variance is Var(X) = E(X²) - [E(X)]², which measures the spread or dispersion of the distribution. A larger variance means the values are more spread out from the mean. The standard deviation is SD(X) = sqrt(Var(X)).
4. Find the mean and variance of a random variable X with P(X = 1) = 0.4, P(X = 2) = 0.3, P(X = 3) = 0.2, P(X = 4) = 0.1.
First verify: 0.4 + 0.3 + 0.2 + 0.1 = 1. Mean: E(X) = 1(0.4) + 2(0.3) + 3(0.2) + 4(0.1) = 0.4 + 0.6 + 0.6 + 0.4 = 2.0. E(X²) = 1(0.4) + 4(0.3) + 9(0.2) + 16(0.1) = 0.4 + 1.2 + 1.8 + 1.6 = 5.0. Variance: Var(X) = E(X²) - [E(X)]² = 5.0 - 4.0 = 1.0.
5. If X follows B(5, 0.2), find P(X = 0) and P(X >= 1).
P(X = 0) = C(5,0)(0.2)⁰(0.8)⁵ = 1(1)(0.32768) = 0.32768. P(X >= 1) = 1 - P(X = 0) = 1 - 0.32768 = 0.67232. Finding P(X >= 1) by using 1 - P(X = 0) is much faster than computing all five individual probabilities.
VI. Answer in Detail (5-Mark Questions)
1. Define a binomial distribution and derive its mean and variance. Illustrate with an example.
A binomial distribution describes the number of successes in n independent Bernoulli trials, each with probability of success p. The probability mass function is P(X = r) = nCr p^r q^(n-r) for r = 0, 1, 2, ..., n, where q = 1 - p. Derivation of mean: E(X) = sum of r times nCr p^r q^(n-r) from r=0 to n. Using the identity and properties of the binomial series, E(X) = np. Derivation of variance: E(X²) = npq + n²p². Therefore Var(X) = E(X²) - [E(X)]² = npq + n²p² - n²p² = npq. Example: A coin is tossed 6 times. Find the probability of getting exactly 4 heads. Here n = 6, p = 1/2, q = 1/2. P(X = 4) = C(6,4)(1/2)⁴(1/2)² = 15(1/16)(1/4) = 15/64. The mean is np = 6(1/2) = 3 and variance is npq = 6(1/2)(1/2) = 3/2.
2. A random variable X has the following probability distribution: P(X = 0) = k, P(X = 1) = 2k, P(X = 2) = 3k, P(X = 3) = k. Find k, E(X), and Var(X).
First, find k using the condition that all probabilities sum to 1: k + 2k + 3k + k = 7k = 1, so k = 1/7. The probability distribution is: P(0) = 1/7, P(1) = 2/7, P(2) = 3/7, P(3) = 1/7. Mean: E(X) = 0(1/7) + 1(2/7) + 2(3/7) + 3(1/7) = 0 + 2/7 + 6/7 + 3/7 = 11/7. E(X²) = 0(1/7) + 1(2/7) + 4(3/7) + 9(1/7) = 0 + 2/7 + 12/7 + 9/7 = 23/7. Variance: Var(X) = E(X²) - [E(X)]² = 23/7 - (11/7)² = 23/7 - 121/49 = 161/49 - 121/49 = 40/49. Standard deviation = sqrt(40/49) = 2sqrt(10)/7.
3. Explain the Poisson distribution and state its properties. If lambda = 2, find P(X = 0), P(X = 1), and P(X = 2).
The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space, when these events occur with a known constant mean rate and independently of the time since the last event. Its probability mass function is P(X = x) = (e^(-lambda) lambda^x) / x! for x = 0, 1, 2, ... . Properties: (i) Both mean and variance equal lambda, (ii) it is a limiting case of the binomial distribution when n is large and p is small with np = lambda, (iii) the distribution is defined for non-negative integers only, (iv) it is skewed to the right. For lambda = 2: P(X = 0) = e^(-2)(2⁰)/0! = e^(-2) = 0.1353. P(X = 1) = e^(-2)(2¹)/1! = 2e^(-2) = 0.2707. P(X = 2) = e^(-2)(2²)/2! = 4e^(-2)/2 = 2e^(-2) = 0.2707. Notice that P(X = 1) = P(X = 2) when lambda = 2.
4. In a binomial distribution B(n, p), if the mean is 6 and the variance is 4, find n, p, and P(X = 3).
Common Mistakes to Avoid
1. ❌ Forgetting to verify that the sum of all probabilities equals 1. ✅ Always check that sum of P(X = x) = 1 before computing mean or variance.
2. ❌ Computing variance as E(X²) instead of E(X²) - [E(X)]². ✅ Variance = E(X²) - [E(X)]². The subtraction of [E(X)]² is essential.
3. ❌ Confusing p and q in binomial distribution. ✅ p is the probability of success and q = 1 - p is the probability of failure. Always compute q from p.
4. ❌ Using the wrong value of nCr in binomial probability. ✅ nCr = n! / (r!(n-r)!). Double-check the combination calculation. Common values: C(5,2) = 10, C(6,3) = 20, C(10,3) = 120.
5. ❌ Forgetting that Poisson mean and variance are both equal to lambda. ✅ In Poisson distribution, E(X) = lambda and Var(X) = lambda. This is a unique property.
6. ❌ Not using 1 - P(X = 0) shortcut for "at least one" probability. ✅ P(X >= 1) = 1 - P(X = 0) is much faster than summing all individual probabilities.
7. ❌ Writing P(X = r) = nCr p^r q^(n-r) with incorrect exponents. ✅ The exponent of p is r (number of successes) and the exponent of q is n-r (number of failures). They must sum to n.
8. ❌ Assuming all probability distributions are binomial. ✅ Check all four conditions (fixed trials, two outcomes, constant p, independence) before applying the binomial formula.
Exam Checklist
- Write the probability distribution table clearly with all values of X and P(X = x).
- Verify that the sum of all probabilities equals 1 before proceeding.
- Show the formula for mean: E(X) = sum of x times P(X = x).
- Show the formula for variance: Var(X) = E(X²) - [E(X)]².
- For binomial distribution, clearly state n, p, and q = 1 - p.
- Use the binomial formula P(X = r) = nCr p^r q^(n-r) with correct exponents.
- For "at least one" problems, use P(X >= 1) = 1 - P(X = 0).
- For Poisson distribution, identify lambda and use P(X = x) = e^(-lambda) lambda^x / x!.
- Simplify final probabilities as fractions or decimals as required.
- State whether the answer is exact or approximate when using e^(-lambda).
- Double-check combination values nCr before computing binomial probabilities.