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Statistics and Probability - Discrete random variables

Grade 11IB_AI

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 (e.g., 0,1,2,…0, 1, 2, \dots).

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The probability distribution of XX describes the probability P(X=x)P(X = x) for each possible value xx.

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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 ∑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 average outcome if the experiment is repeated many times.

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The Variance Var(X)Var(X) measures the spread of the distribution, and the standard deviation is given by σ=Var(X)\sigma = \sqrt{Var(X)}.

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

📐Formulae

∑P(X=x)=1\sum P(X = x) = 1

E(X)=μ=∑x⋅P(X=x)E(X) = \mu = \sum x \cdot P(X = x),

Var(X)=∑(x−μ)2P(X=x)=E(X2)−[E(X)]2Var(X) = \sum (x - \mu)^2 P(X = x) = E(X^2) - [E(X)]^2

P(X=r)=(nr)pr(1−p)n−rP(X = r) = \binom{n}{r} p^r (1 - p)^{n - r}

E(X)=npandVar(X)=np(1−p)E(X) = np \quad \text{and} \quad Var(X) = np(1 - p)

💡Examples

Problem 1:

The discrete random variable XX has the following probability distribution: x1234P(X=x)0.1k0.32k\begin{array}{|c|c|c|c|c|} \hline x & 1 & 2 & 3 & 4 \\ \hline P(X=x) & 0.1 & k & 0.3 & 2k \\ \hline \end{array} Find the value of kk and calculate E(X)E(X).

Solution:

  1. Since the sum of probabilities must be 11: 0.1+k+0.3+2k=10.1 + k + 0.3 + 2k = 1 0.4+3k=10.4 + 3k = 1 3k=0.6  ⟹  k=0.23k = 0.6 \implies k = 0.2

  2. Now, substitute kk back into the table: P(X=2)=0.2P(X=2) = 0.2 and P(X=4)=0.4P(X=4) = 0.4.

  3. Calculate E(X)E(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 first use the property that the sum of all probabilities in a distribution equals 11 to solve for the unknown constant kk. Once kk is found, we apply the formula for the expected value by summing the product of each value and its corresponding probability.

Problem 2:

A fair six-sided die is rolled 1010 times. Let XX be the number of times a '6' is rolled. Find the probability of rolling exactly three 6s.

Solution:

This follows a Binomial Distribution X∼B(n,p)X \sim B(n, p) where n=10n = 10 and p=16p = \frac{1}{6}. We want to find P(X=3)P(X = 3): P(X=3)=(103)(16)3(56)10−3P(X = 3) = \binom{10}{3} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^{10-3} P(X=3)=120×(1216)×(78125279936)≈0.155P(X = 3) = 120 \times \left(\frac{1}{216}\right) \times \left(\frac{78125}{279936}\right) \approx 0.155

Explanation:

Since the trials are independent and there is a constant probability of success (1/61/6), we use the Binomial Probability Mass Function. In IB AI, this can also be calculated using the 'Binomial PDF' function on a GDC (Graphic Display Calculator).