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Statistics and Probability - Distributions – Discrete random variables

Grade 11IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Discrete Random Variable XX is a variable that can take on a countable number of distinct values. The probability that XX takes a specific value xx is denoted by P(X=x)P(X = x).

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For any discrete probability distribution, two conditions must be met: 0≤P(X=x)≤10 \le P(X = x) \le 1 for all xx, and the sum of all probabilities must be equal to 11, expressed as ∑P(X=x)=1\sum P(X = x) = 1.

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The Expected Value E(X)E(X), also known as the mean μ\mu, represents the long-term average of the outcomes: E(X)=∑x⋅P(X=x)E(X) = \sum x \cdot P(X = x).

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The Variance Var(X)Var(X) measures the spread of the distribution: Var(X)=E(X2)−[E(X)]2Var(X) = E(X^2) - [E(X)]^2. The standard deviation σ\sigma is the square root of the variance, σ=Var(X)\sigma = \sqrt{Var(X)}.

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A Binomial Distribution X∼B(n,p)X \sim B(n, p) occurs when there are nn independent trials, each with a constant probability of success pp and only two possible outcomes (success or failure).

📐Formulae

∑i=1nP(X=xi)=1\sum_{i=1}^{n} P(X = x_i) = 1

E(X)=μ=∑xiP(X=xi)E(X) = \mu = \sum x_i P(X = x_i)

Var(X)=σ2=E(X2)−μ2=∑xi2P(X=xi)−μ2Var(X) = \sigma^2 = E(X^2) - \mu^2 = \sum x_i^2 P(X = x_i) - \mu^2

P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

E(X)=np (for Binomial Distribution)E(X) = np \text{ (for Binomial Distribution)}

Var(X)=np(1−p) (for Binomial Distribution)Var(X) = np(1-p) \text{ (for Binomial Distribution)}

💡Examples

Problem 1:

The discrete random variable XX has the following probability distribution: P(X=1)=0.1P(X=1) = 0.1, P(X=2)=kP(X=2) = k, P(X=3)=0.3P(X=3) = 0.3, and P(X=4)=2kP(X=4) = 2k. Find the value of kk and calculate E(X)E(X).

Solution:

First, use the property ∑P(X=x)=1\sum P(X=x) = 1: 0.1+k+0.3+2k=10.1 + k + 0.3 + 2k = 1 3k+0.4=13k + 0.4 = 1 3k=0.6  ⟹  k=0.23k = 0.6 \implies k = 0.2 Now, calculate E(X)E(X) using E(X)=∑x⋅P(X=x)E(X) = \sum x \cdot P(X=x): E(X)=(1×0.1)+(2×0.2)+(3×0.3)+(4×0.4)E(X) = (1 \times 0.1) + (2 \times 0.2) + (3 \times 0.3) + (4 \times 0.4) E(X)=0.1+0.4+0.9+1.6=3.0E(X) = 0.1 + 0.4 + 0.9 + 1.6 = 3.0

Explanation:

We sum all probabilities to solve for the unknown constant kk. Once kk is found, we multiply each outcome xx by its probability and sum them to find the mean.

Problem 2:

A fair die is rolled 1212 times. Let XX be the number of times a '66' is rolled. Find P(X=2)P(X = 2).

Solution:

This is a binomial distribution X∼B(n,p)X \sim B(n, p) where n=12n = 12 and p=16p = \frac{1}{6}. We want to find P(X=2)P(X = 2): P(X=2)=(122)(16)2(56)10P(X = 2) = \binom{12}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^{10} (122)=12×112×1=66\binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66 P(X=2)=66×(136)×(56)10≈0.296P(X = 2) = 66 \times \left(\frac{1}{36}\right) \times \left(\frac{5}{6}\right)^{10} \approx 0.296

Explanation:

The problem fits the binomial criteria: fixed number of trials (1212), independent outcomes, and constant probability of success (rolling a '66').

Problem 3:

In a game, a player wins Rs 10 if they roll a '66' on a fair die and loses Rs 2 for any other number. Calculate the expected profit per game.

Solution:

Let XX be the profit. P(X=10)=16P(X = 10) = \frac{1}{6} P(X=−2)=56P(X = -2) = \frac{5}{6} E(X)=(10×16)+(−2×56)E(X) = (10 \times \frac{1}{6}) + (-2 \times \frac{5}{6}) E(X)=106−106=0E(X) = \frac{10}{6} - \frac{10}{6} = 0 Alternatively, using the array format for simple calculation: 10/6−10/60\begin{array}{r} 10/6 \\ -10/6 \\ \hline 0 \end{array}

Explanation:

The expected profit is calculated by multiplying each monetary outcome by its probability. An expected value of 00 indicates a 'fair game'.

Distributions – Discrete random variables Grade 11 Notes & Examples