Grade 12

Advanced

7 min

Lesson 11 of 12

Advanced Probability Distributions

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Analyze discrete and continuous random variables using binomial and Poisson distributions.

If you flip a coin 100 times, you expect 50 heads—but how 'surprised' should you be if you actually get 60? Probability distributions allow us to quantify that surprise and predict the long-term behavior of chaotic systems.

Learning Objectives

1.Calculate the expected value and variance for discrete random variables to determine long-term averages.
2.Apply the Binomial Distribution formula to solve multi-trial probability problems in real-world contexts.
3.Differentiate between discrete Probability Mass Functions (PMF) and continuous Probability Density Functions (PDF).

The Heart of Discrete Variables

A discrete random variable is a variable that can take on a countable number of distinct values. To understand its behavior, we look at two primary metrics: Expected Value

E(X)

and Variance

Var(X)

. The Expected Value is essentially the 'long-term average' of the variable. We calculate it by summing the product of each possible value and its probability:

E(X) = \sum x \cdot P(X=x)

Variance measures the 'spread' or risk of the distribution. It tells us how much the outcomes typically deviate from the mean. The most efficient formula for variance is:

Var(X) = E(X^2) - [E(X)]^2

where

E(X^2) = \sum x^2 \cdot P(X=x)

Quick Check

If a random variable X can be 0 with probability 0.4 and 10 with probability 0.6, what is the Expected Value E(X)?

Multiply each value by its probability and add them together: (0 0.4) + (10 0.6).

Answer

The Binomial Powerhouse

The Binomial Distribution is used when we have a fixed number of independent trials,

n

, each with only two possible outcomes: 'success' or 'failure'. The probability of success,

p

, must remain constant across all trials. We use the binomial formula to find the probability of getting exactly

k

successes:

P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}

Here,

\binom{n}{k}

is the binomial coefficient, calculated as

\frac{n!}{k!(n-k)!}

. This distribution is foundational for quality control, medical testing, and even sports analytics.

A basketball player has an 80% chance ( $p=0.8$ ) of making a free throw. If they take 3 shots ( $n=3$ ), what is the probability they make exactly 2 ( $k=2$ )?

1. Identify variables: $n=3, k=2, p=0.8, (1-p)=0.2$ . 2. Set up the formula: $P(X=2) = \binom{3}{2} (0.8)^2 (0.2)^1$ . 3. Calculate the coefficient: $\binom{3}{2} = 3$ . 4. Solve: $3 \cdot 0.64 \cdot 0.2 = 0.384$ . 5. Result: There is a 38.4% chance of making exactly 2 shots.

Quick Check

In a Binomial Distribution, if the probability of success p is 0.5, what is the probability of failure (1-p)?

The sum of the probabilities of success and failure must always equal 1.

Answer

0.5

Discrete vs. Continuous Landscapes

While discrete variables deal with 'counts' (like the number of heads), continuous random variables deal with 'measurements' (like height or time). For discrete variables, we use a Probability Mass Function (PMF) where

P(X=x)

is a specific value. However, for continuous variables, the probability of any exact point is zero! Instead, we use a Probability Density Function (PDF), denoted as

f(x)

. The probability is found by calculating the area under the curve over an interval using integration:

P(a \le X \le b) = \int_{a}^{b} f(x) dx

The total area under any valid PDF must always equal 1.

The Poisson Distribution is a discrete distribution often used for the number of events occurring in a fixed interval of time. If a cafe receives an average of $\lambda = 4$ customers per hour, what is the probability of exactly 2 customers arriving in the next hour?

1. Formula: $P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}$ . 2. Plug in values: $P(X=2) = \frac{4^2 e^{-4}}{2!}$ . 3. Simplify: $\frac{16 \cdot 0.0183}{2} = 8 \cdot 0.0183$ . 4. Result: $P(X=2) \approx 0.1465$ or 14.65%.

Prove that for a Binomial Distribution, the variance is $np(1-p)$ . Suppose $n=100$ and $p=0.2$ .

1. Calculate $E(X) = np = 100 \cdot 0.2 = 20$ . 2. Calculate $Var(X) = np(1-p) = 100 \cdot 0.2 \cdot 0.8 = 16$ . 3. Standard Deviation: $\sigma = \sqrt{Var(X)} = 4$ . 4. Interpretation: In 100 trials, you expect 20 successes, but most results will fall between $20 \pm 4$ (16 to 24 successes).

Key Takeaways

1The Expected Value $E(X)$ represents the long-term theoretical average of a random variable.
2The Binomial Distribution requires independent trials, a fixed $n$ , and a constant probability $p$ .
3Continuous distributions (PDFs) measure probability as the area under a curve, not as discrete points.
4Variance $Var(X)$ quantifies the spread of outcomes, where higher variance indicates higher uncertainty.

Quick Check

Multiple Choice

Which of the following is a requirement for a Binomial Distribution?

Multiple Choice

If the total area under a Probability Density Function (PDF) is calculated, what must it equal?

True/False

The Expected Value E(X) of a random variable is always one of the possible outcomes of the variable.

What's Next?

Review Tomorrow

In 24 hours, try to write down the Binomial probability formula from memory and explain what each variable (n, k, p) represents.

Practice Activity

Find a real-world 'success/failure' scenario (like a sports stat or a manufacturing defect rate) and calculate the probability of 3 successes in 10 trials using the formula.

Browse

Browse

Advanced Probability Distributions

Learning Objectives

The Heart of Discrete Variables

The Binomial Powerhouse

Discrete vs. Continuous Landscapes

Key Takeaways

Quick Check

What's Next?

Advanced Probability Distributions

Learning Objectives

The Heart of Discrete Variables

The Binomial Powerhouse

Discrete vs. Continuous Landscapes

Key Takeaways

Quick Check

What's Next?