krit.club logo

Probability - Random Variables and Probability Distributions

Grade 12ICSE

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

🔑Concepts

•

A Discrete Random Variable is a real-valued function defined on the sample space of a random experiment. Visually, it maps every outcome (like 'Heads' or 'Tails') to a specific point on the real number line, transforming qualitative outcomes into quantitative data.

•

The Probability Distribution of a discrete random variable is a comprehensive list or table of all possible values xix_i and their associated probabilities P(X=xi)P(X = x_i). Visually, this can be represented as a probability histogram or bar chart where the x-axis represents the values of the random variable and the y-axis represents the probabilities; the sum of the heights of all bars must equal 1.

•

The Mean or Expected Value, denoted by E(X)E(X) or μ\mu, is the long-term average outcome of the random variable. In a visual distribution plot, the mean represents the 'center of mass' or the 'balancing point' where the distribution would remain level if placed on a pivot.

•

Variance (Var(X)Var(X)) and Standard Deviation (σ\sigma) quantify the spread or dispersion of the random variable around its mean. Visually, a small standard deviation results in a distribution graph that is tall and narrow (highly clustered), while a large standard deviation results in a graph that is short and widely spread out.

•

The Cumulative Distribution Function (CDF), denoted as F(x)=P(X≤x)F(x) = P(X \le x), provides the probability that XX will take a value less than or equal to xx. Visually, the CDF is a non-decreasing 'staircase' or step function that starts at height 0 on the left and reaches height 1 on the right.

•

Bernoulli Trials are independent experiments with exactly two possible outcomes: 'Success' and 'Failure'. A sequence of nn such trials leads to the Binomial Distribution. Visually, if the probability of success p=0.5p = 0.5, the binomial distribution is perfectly symmetrical (bell-shaped); if p<0.5p < 0.5, the distribution is skewed to the right.

•

The total probability for any discrete distribution must satisfy two conditions: each individual probability pi≥0p_i \ge 0 and the sum of all probabilities ∑pi=1\sum p_i = 1. This ensures that the entire sample space is accounted for.

📐Formulae

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

Mean (Expected Value): E(X)=μ=∑xiP(X=xi)E(X) = \mu = \sum x_i P(X=x_i)

Variance: Var(X)=σ2=E(X2)−[E(X)]2Var(X) = \sigma^2 = E(X^2) - [E(X)]^2

Alternative Variance Form: Var(X)=∑(xi−μ)2P(X=xi)Var(X) = \sum (x_i - \mu)^2 P(X=x_i)

Expectation of X2X^2: E(X2)=∑xi2P(X=xi)E(X^2) = \sum x_i^2 P(X=x_i)

Standard Deviation: σ=Var(X)\sigma = \sqrt{Var(X)}

Binomial Distribution Probability: P(X=r)=(nr)prqn−rP(X = r) = \binom{n}{r} p^r q^{n-r}, where q=1−pq = 1 - p

Mean of Binomial Distribution: μ=np\mu = np

Variance of Binomial Distribution: σ2=npq\sigma^2 = npq

💡Examples

Problem 1:

A random variable XX has the following probability distribution: XX: 0,1,2,30, 1, 2, 3 P(X)P(X): 0.1,k,2k,0.30.1, k, 2k, 0.3 Find the value of kk and calculate the mean E(X)E(X).

Solution:

  1. Using the property that the sum of probabilities is 1: 0.1+k+2k+0.3=10.1 + k + 2k + 0.3 = 1 3k+0.4=13k + 0.4 = 1 3k=0.6  ⟹  k=0.23k = 0.6 \implies k = 0.2

  2. The updated distribution is: X=0,P(X)=0.1X=0, P(X)=0.1 X=1,P(X)=0.2X=1, P(X)=0.2 X=2,P(X)=0.4X=2, P(X)=0.4 X=3,P(X)=0.3X=3, P(X)=0.3

  3. Calculate Mean E(X)E(X): E(X)=∑xiP(xi)=(0×0.1)+(1×0.2)+(2×0.4)+(3×0.3)E(X) = \sum x_i P(x_i) = (0 \times 0.1) + (1 \times 0.2) + (2 \times 0.4) + (3 \times 0.3) E(X)=0+0.2+0.8+0.9=1.9E(X) = 0 + 0.2 + 0.8 + 0.9 = 1.9

Explanation:

We first apply the total probability rule to solve for the unknown constant kk. Once kk is found, we use the definition of Expected Value by multiplying each value of XX by its probability and summing the results.

Problem 2:

A fair die is tossed 5 times. Find the probability of getting exactly 2 sixes.

Solution:

  1. Identify the type of distribution: This is a Binomial Distribution problem since there are nn independent trials with two outcomes (getting a six or not).

  2. Define parameters: Number of trials n=5n = 5 Probability of success (getting a six) p=16p = \frac{1}{6} Probability of failure q=1−p=56q = 1 - p = \frac{5}{6} Number of successes required r=2r = 2

  3. Apply Binomial formula P(X=r)=(nr)prqn−rP(X=r) = \binom{n}{r} p^r q^{n-r}: P(X=2)=(52)(16)2(56)5−2P(X=2) = \binom{5}{2} (\frac{1}{6})^2 (\frac{5}{6})^{5-2} P(X=2)=10×136×(56)3P(X=2) = 10 \times \frac{1}{36} \times (\frac{5}{6})^3 P(X=2)=10×136×125216P(X=2) = 10 \times \frac{1}{36} \times \frac{125}{216} P(X=2)=12507776≈0.1608P(X=2) = \frac{1250}{7776} \approx 0.1608

Explanation:

By identifying the experiment as a series of Bernoulli trials, we use the Binomial formula to calculate the probability of a specific number of successes. The combination factor (52)\binom{5}{2} accounts for the different orders in which the 2 sixes can occur.