Probability: Random Variables, Mean & Variance (Ex 13.4, Class 12 Maths NCERT)

Understanding Random Variables and Probability Distributions

Imagine you toss two coins. The possible outcomes are HH, HT, TH, TT. If we're interested in the 'number of heads', we can assign a numerical value to each outcome: 2 for HH, 1 for HT, 1 for TH, and 0 for TT. This numerical value, whose value is determined by the outcome of a random experiment, is called a Random Variable. We usually denote it by capital letters like X, Y, Z.

Types of Random Variables

1. Discrete Random Variable: A random variable is discrete if its possible values are finite or countably infinite (like the number of heads in coin tosses, number of defective items, etc.). These values can often be listed.
2. Continuous Random Variable: A random variable is continuous if it can take any value within a given interval (like height, weight, time taken to complete a task). This exercise primarily focuses on discrete random variables.

Probability Distribution of a Discrete Random Variable

The probability distribution of a discrete random variable X is a list or table that shows all possible values 'x_i' that X can take, along with their corresponding probabilities P(X=x_i). These probabilities are often called the Probability Mass Function (PMF), denoted as P(x_i).

For a valid probability distribution, two conditions must be met:

1. P(X = x_i) ≥ 0 for all possible values x_i (probabilities cannot be negative).
2. Σ P(X = x_i) = 1 (the sum of all probabilities must be equal to 1, as one of the outcomes must occur).

Mean (Expectation) of a Random Variable

Mean (Expectation), E(X) or μ

The mean or expectation of a discrete random variable X, denoted as E(X) or μ (mu), is a measure of the central tendency of its probability distribution. It represents the average value of the random variable if the experiment is repeated many times. For a discrete random variable X with values x₁, x₂, ..., xₙ and corresponding probabilities P(x₁), P(x₂), ..., P(xₙ), the mean is calculated as: **E(X) = Σ (x_i * P(X = x_i))**

Variance and Standard Deviation of a Random Variable

Variance, Var(X) or σ²

The variance of a discrete random variable X, denoted as Var(X) or σ² (sigma squared), measures the spread or dispersion of the probability distribution around its mean. A higher variance indicates that the values of the random variable are more spread out from the mean. It is calculated using two equivalent formulas: 1. Var(X) = E(X²) - [E(X)]² where E(X²) = Σ (x_i² * P(X = x_i)) 2. **Var(X) = Σ ((x_i - μ)² * P(X = x_i))** Both formulas yield the same result, but the first one is often computationally easier.

Standard Deviation, SD(X) or σ

The standard deviation of a random variable X, denoted as SD(X) or σ, is the positive square root of its variance. It provides a measure of spread in the same units as the random variable itself, making it more interpretable than variance. SD(X) = √Var(X) = √σ² = σ

Worked Examples: Applying Probability Distribution Concepts

• Example 1: Finding Probability Distribution A bag contains 2 red and 3 black balls. Two balls are drawn at random without replacement. Let X be the random variable representing the number of black balls drawn. Find the probability distribution of X. Solution: Step 1: Identify all possible outcomes and values of X. Total balls = 2 (Red) + 3 (Black) = 5 balls. We draw 2 balls. The number of black balls (X) can be 0, 1, or 2. Step 2: Calculate the probability for each value of X. P(X=0): Probability of drawing 0 black balls (i.e., 2 red balls). P(X=0) = (Number of ways to choose 0 black balls from 3) (Number of ways to choose 2 red balls from 2) / (Total ways to choose 2 balls from 5) P(X=0) = (³C₀ ²C₂) / ⁵C₂ = (1 1) / 10 = 1/10 P(X=1): Probability of drawing 1 black ball and 1 red ball. P(X=1) = (³C₁ ²C₁) / ⁵C₂ = (3 2) / 10 = 6/10 = 3/5 P(X=2): Probability of drawing 2 black balls (i.e., 0 red balls). P(X=2) = (³C₂ ²C₀) / ⁵C₂ = (3 1) / 10 = 3/10 Step 3: Create the probability distribution table. | X (Number of Black Balls) | P(X) | | :------------------------ | :--- | | 0 | 1/10 | | 1 | 6/10 | | 2 | 3/10 | Step 4: Verify the conditions. All P(X) values are ≥ 0. Sum P(X) = 1/10 + 6/10 + 3/10 = 10/10 = 1. The distribution is valid. Final Answer: The probability distribution of X is: | X | 0 | 1 | 2 | | ----- | --- | --- | --- | | P(X) | 1/10 | 6/10 | 3/10 |
• Example 2: Calculating Mean and Variance For the following probability distribution of a random variable X, find the mean and variance: | X | 0 | 1 | 2 | | :---- | :---- | :---- | :---- | | P(X) | 0.2 | 0.5 | 0.3 | Solution: Step 1: Calculate the Mean (E(X)). E(X) = Σ (x_i P(X = x_i)) E(X) = (0 0.2) + (1 0.5) + (2 0.3) E(X) = 0 + 0.5 + 0.6 = 1.1 Step 2: Calculate E(X²). E(X²) = Σ (x_i² P(X = x_i)) E(X²) = (0² 0.2) + (1² 0.5) + (2² 0.3) E(X²) = (0 0.2) + (1 0.5) + (4 * 0.3) E(X²) = 0 + 0.5 + 1.2 = 1.7 Step 3: Calculate the Variance (Var(X)). Var(X) = E(X²) - [E(X)]² Var(X) = 1.7 - (1.1)² Var(X) = 1.7 - 1.21 = 0.49 Step 4: Calculate the Standard Deviation (SD(X)). SD(X) = √Var(X) = √0.49 = 0.7 Final Answer: The mean (E(X)) is 1.1, the variance (Var(X)) is 0.49, and the standard deviation (SD(X)) is 0.7.

Exam Tips and Common Mistakes to Avoid

To ace problems from Exercise 13.4, keep these points in mind:

• Verify Probability Distribution Properties: Always cross-check that all P(X=xᵢ) values are non-negative and their sum equals 1. This is a common point for error and a quick check can save marks.
• Distinguish X from xᵢ: Remember, X is the random variable (the function), while xᵢ represents its specific possible values (the outcomes). This conceptual clarity is vital.
• Careful with Combinations/Permutations: When setting up the probability distribution, especially for drawing balls, cards, or dice rolls, ensure you correctly use combinations (nCr) or permutations (nPr) as required by the problem's context (e.g., 'without replacement' usually implies combinations for simultaneous draws).
• Formula Accuracy for Variance: There are two formulas for variance. While Var(X) = E(X²) - [E(X)]² is often simpler, ensure you correctly calculate E(X²) by summing xᵢ²P(xᵢ) – not just squaring E(X) directly.
• Step-by-Step Calculation: For mean and variance, creating a tabular format (xᵢ, P(xᵢ), xᵢP(xᵢ), xᵢ²P(xᵢ)) can help organize your calculations and reduce arithmetic errors.

Practice Questions with Solutions

• Q: An urn contains 4 white and 6 red balls. Three balls are drawn one by one without replacement. If X denotes the number of white balls drawn, find the probability distribution of X. A: Step 1: Identify possible values for X. Total balls = 4 white + 6 red = 10 balls. We draw 3 balls. X, the number of white balls, can be 0, 1, 2, or 3. Step 2: Calculate probabilities for each X. Total ways to draw 3 balls from 10 = ¹⁰C₃ = (10 9 8) / (3 2 1) = 120. P(X=0): (⁴C₀ ⁶C₃) / ¹⁰C₃ = (1 (654)/(321)) / 120 = (1 20) / 120 = 20/120 = 1/6. P(X=1): (⁴C₁ ⁶C₂) / ¹⁰C₃ = (4 (65)/2) / 120 = (4 15) / 120 = 60/120 = 1/2. P(X=2): (⁴C₂ ⁶C₁) / ¹⁰C₃ = ((43)/2 6) / 120 = (6 6) / 120 = 36/120 = 3/10. P(X=3): (⁴C₃ ⁶C₀) / ¹⁰C₃ = (4 * 1) / 120 = 4/120 = 1/30. Step 3: Form the probability distribution table. | X | 0 | 1 | 2 | 3 | | ----- | ----- | ----- | ----- | ----- | | P(X) | 1/6 | 1/2 | 3/10 | 1/30 | Final answer: The probability distribution of X is: | X | 0 | 1 | 2 | 3 | | ----- | ----- | ----- | ----- | ----- | | P(X) | 1/6 | 1/2 | 3/10 | 1/30 |
• Q: A die is tossed twice. Let X be the number of times a number greater than 4 appears. Find the mean and variance of X. A: Step 1: Define the random variable and its possible values. A number greater than 4 means 5 or 6. Probability of getting > 4 in one toss = 2/6 = 1/3. Probability of not getting > 4 = 4/6 = 2/3. X is the number of times a number greater than 4 appears in two tosses. X can take values 0, 1, or 2. Step 2: Calculate the probability distribution for X. P(X=0): (Not > 4 on 1st toss) AND (Not > 4 on 2nd toss) = (2/3) (2/3) = 4/9. P(X=1): ('> 4' on 1st, 'not > 4' on 2nd) OR ('not > 4' on 1st, '> 4' on 2nd) = (1/3 2/3) + (2/3 1/3) = 2/9 + 2/9 = 4/9. P(X=2): ('> 4' on 1st toss) AND ('> 4' on 2nd toss) = (1/3) (1/3) = 1/9. Probability Distribution: | X | 0 | 1 | 2 | | ----- | ----- | ----- | ----- | | P(X) | 4/9 | 4/9 | 1/9 | (Check: 4/9 + 4/9 + 1/9 = 9/9 = 1) Step 3: Calculate E(X). E(X) = (0 4/9) + (1 4/9) + (2 1/9) = 0 + 4/9 + 2/9 = 6/9 = 2/3. Step 4: Calculate E(X²). E(X²) = (0² 4/9) + (1² 4/9) + (2² 1/9) = (0 4/9) + (1 4/9) + (4 1/9) = 0 + 4/9 + 4/9 = 8/9. Step 5: Calculate Var(X). Var(X) = E(X²) - [E(X)]² = 8/9 - (2/3)² = 8/9 - 4/9 = 4/9. Final answer: The mean (E(X)) is 2/3 and the variance (Var(X)) is 4/9.
• Q: A random variable X has the following probability distribution: | X | -1 | 0 | 1 | 2 | | :---- | :---- | :---- | :---- | :---- | | P(X) | k | 0.2 | 0.3 | 0.1 | Find the value of k and then find E(X). A: Step 1: Use the property that the sum of probabilities is 1 to find k. Σ P(X = x_i) = 1 k + 0.2 + 0.3 + 0.1 = 1 k + 0.6 = 1 k = 1 - 0.6 = 0.4 Step 2: Calculate E(X) using the value of k. E(X) = Σ (x_i P(X = x_i)) E(X) = (-1 k) + (0 0.2) + (1 0.3) + (2 0.1) E(X) = (-1 0.4) + 0 + 0.3 + 0.2 E(X) = -0.4 + 0.5 = 0.1 Final answer: The value of k is 0.4, and the mean E(X) is 0.1.
• Q: Find the standard deviation for the random variable X with the following probability distribution: | X | 1 | 2 | 3 | | :---- | :---- | :---- | :---- | | P(X) | 0.1 | 0.6 | 0.3 | A: Step 1: Calculate E(X). E(X) = Σ (x_i P(X = x_i)) E(X) = (1 0.1) + (2 0.6) + (3 0.3) E(X) = 0.1 + 1.2 + 0.9 = 2.2 Step 2: Calculate E(X²). E(X²) = Σ (x_i² P(X = x_i)) E(X²) = (1² 0.1) + (2² 0.6) + (3² 0.3) E(X²) = (1 0.1) + (4 0.6) + (9 * 0.3) E(X²) = 0.1 + 2.4 + 2.7 = 5.2 Step 3: Calculate Var(X). Var(X) = E(X²) - [E(X)]² Var(X) = 5.2 - (2.2)² Var(X) = 5.2 - 4.84 = 0.36 Step 4: Calculate SD(X). SD(X) = √Var(X) = √0.36 = 0.6 Final answer: The standard deviation of X is 0.6.

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