Solved Exercises on Probability Distributions Introduction in Grade 10

Master probability distributions: discrete and continuous distributions, probability mass functions, and cumulative distribution functions through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Discrete Probability Distribution
Exercise 1
A fair six-sided die is rolled. Let X be the random variable representing the outcome. Find the probability distribution of X and calculate P(X ≤ 3).
Definition:

Discrete Random Variable: Takes on countable number of distinct values

Probability Mass Function (PMF): Function that assigns probability to each possible value

Properties: 1) 0 ≤ P(X = x) ≤ 1, 2) Σ P(X = x) = 1

Discrete distribution method:
  1. Identify all possible outcomes of the random variable
  2. Calculate probability for each outcome
  3. Verify probabilities sum to 1
  4. For cumulative probability, sum individual probabilities
Possible values
X ∈ {1, 2, 3, 4, 5, 6}
PMF
P(X = x) = 1/6
CDF
P(X ≤ 3)
Step 1: Identify possible outcomes

Die can show values: 1, 2, 3, 4, 5, 6

Step 2: Calculate probability for each outcome

Since die is fair: P(X = 1) = P(X = 2) = ... = P(X = 6) = 1/6

Step 3: Verify probability sum

P(X = 1) + P(X = 2) + ... + P(X = 6) = 6 × (1/6) = 1 ✓

Step 4: Calculate cumulative probability

P(X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3) = 1/6 + 1/6 + 1/6 = 3/6 = 1/2

P(X ≤ 3) = 1/2
Final answer:

The probability distribution of X is uniform with P(X = x) = 1/6 for x = 1,2,3,4,5,6, and P(X ≤ 3) = 1/2

Applied rules:

• Probability axioms: Probabilities are between 0 and 1

• Normalization: Sum of all probabilities equals 1

• Cumulative probability: P(X ≤ k) = Σ P(X = i) for i ≤ k

2 Binomial Distribution
Exercise 2
A biased coin lands heads with probability 0.3. If the coin is flipped 5 times, find the probability of getting exactly 2 heads.
Definition:

Binomial Distribution: Models number of successes in n independent Bernoulli trials

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

Parameters: n (trials), p (probability of success)

Given
n = 5, k = 2, p = 0.3
Formula
\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)
Result
0.3087
Step 1: Identify parameters

n = 5 (number of flips), k = 2 (number of heads), p = 0.3 (probability of heads)

Step 2: Apply binomial formula

\(P(X = 2) = \binom{5}{2} (0.3)^2 (0.7)^3\)

Step 3: Calculate combination

\(\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{120}{2 \times 6} = 10\)

Step 4: Calculate probabilities

\((0.3)^2 = 0.09\), \((0.7)^3 = 0.343\)

\(P(X = 2) = 10 \times 0.09 \times 0.343 = 0.3087\)

P(X = 2) = 0.3087
Final answer:

The probability of getting exactly 2 heads in 5 flips is 0.3087 or 30.87%

Applied rules:

• Binomial formula: \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

• Combinations: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)

• Independent trials: Each flip is independent

3 Continuous Uniform Distribution
Exercise 3
The waiting time at a bus stop follows a uniform distribution between 0 and 15 minutes. Find the probability that a passenger waits between 5 and 10 minutes.
Definition:

Continuous Random Variable: Takes on uncountably infinite number of values in an interval

Probability Density Function (PDF): Function f(x) such that P(a ≤ X ≤ b) = ∫[a,b] f(x)dx

Uniform PDF: f(x) = 1/(b-a) for a ≤ x ≤ b, 0 otherwise

Given
X ~ U(0, 15), find P(5 ≤ X ≤ 10)
PDF
f(x) = 1/15
Probability
5/15 = 1/3
Step 1: Identify the distribution

X follows uniform distribution over [0, 15], so a = 0, b = 15

Step 2: Calculate PDF

f(x) = 1/(b-a) = 1/(15-0) = 1/15 for 0 ≤ x ≤ 15

Step 3: Calculate probability for uniform distribution

For uniform distribution: P(c ≤ X ≤ d) = (d-c)/(b-a)

P(5 ≤ X ≤ 10) = (10-5)/(15-0) = 5/15 = 1/3

Step 4: Interpret the result

The probability is 1/3 or approximately 33.33%

P(5 ≤ X ≤ 10) = 1/3
Final answer:

The probability that a passenger waits between 5 and 10 minutes is 1/3 or approximately 33.33%

Applied rules:

• Uniform distribution: f(x) = 1/(b-a) for [a,b]

• Probability calculation: Area under PDF curve

• For uniform: P(c ≤ X ≤ d) = (d-c)/(b-a)

Probability Distributions Guide
\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \text{ (Binomial)}, f(x) = \frac{1}{b-a} \text{ (Uniform)}\)
Distribution Formulas
Discrete
PMF: P(X = x)
Examples: Dice rolls, coin flips, counts
Continuous
PDF: f(x)
Examples: Heights, weights, time intervals
Cumulative
CDF: F(x) = P(X ≤ x)
Probability up to a value
Key definitions:

Random Variable: Variable whose value depends on outcome of random experiment

Probability Distribution: Description of possible values and their probabilities

Expected Value: Long-run average value, E[X] = Σx·P(X=x) for discrete

Distribution analysis methodology:
  1. Identify type: Discrete vs continuous
  2. Find parameters: Mean, variance, distribution family
  3. Apply formula: PMF/PDF for specific probabilities
  4. Calculate probabilities: Using appropriate method
Tip 1: For discrete variables, probabilities are point masses; for continuous, they are areas under curves.
Tip 2: In continuous distributions, P(X = a) = 0 for any specific value a.
Tip 3: CDF is always non-decreasing and approaches 1 as x increases.
Tip 4: Check that all probabilities sum to 1 (discrete) or PDF integrates to 1 (continuous).
Common errors: Confusing PMF with PDF, forgetting normalization, mixing discrete and continuous methods.
Exam preparation: Know formulas for common distributions, practice identifying distribution types, work with real-world scenarios.
Essential formulas:

• Discrete: P(X = x) ≥ 0, Σ P(X = x) = 1

• Continuous: f(x) ≥ 0, ∫f(x)dx = 1

• Expected value: E[X] = Σx·P(X=x) (discrete), E[X] = ∫xf(x)dx (continuous)

• Variance: Var(X) = E[X²] - (E[X])²

Solution: Exercises 4 to 5
4 Empirical Probability Distribution
Exercise 4
Based on historical data, the number of accidents per day at an intersection follows this distribution: P(X=0)=0.4, P(X=1)=0.3, P(X=2)=0.2, P(X=3)=0.1. Find the expected number of accidents per day and the probability of having at most 2 accidents.
Definition:

Empirical Distribution: Probability distribution based on observed frequencies

Expected Value: E[X] = Σ x·P(X = x), weighted average of possible outcomes

Cumulative Probability: P(X ≤ k) = Σ P(X = i) for i ≤ k

Given
P(X=0)=0.4, P(X=1)=0.3, P(X=2)=0.2, P(X=3)=0.1
Expected value
E[X] = Σ x·P(X = x)
Cumulative prob
P(X ≤ 2)
Step 1: Verify probability sum

P(X=0) + P(X=1) + P(X=2) + P(X=3) = 0.4 + 0.3 + 0.2 + 0.1 = 1.0 ✓

Step 2: Calculate expected value

E[X] = 0×P(X=0) + 1×P(X=1) + 2×P(X=2) + 3×P(X=3)

E[X] = 0×0.4 + 1×0.3 + 2×0.2 + 3×0.1 = 0 + 0.3 + 0.4 + 0.3 = 1.0

Step 3: Calculate cumulative probability

P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2) = 0.4 + 0.3 + 0.2 = 0.9

Step 4: Interpret results

On average, there is 1 accident per day. There's a 90% chance of having at most 2 accidents.

E[X] = 1.0, P(X ≤ 2) = 0.9
Final answer:

The expected number of accidents per day is 1.0, and the probability of having at most 2 accidents is 0.9 or 90%

Applied rules:

• Expected value formula: E[X] = Σ x·P(X = x)

• Cumulative probability: P(X ≤ k) = Σ P(X = i) for i ≤ k

• Normalization: All probabilities sum to 1

5 Probability Distribution Transformation
Exercise 5
Given that X has the probability distribution P(X=1)=0.2, P(X=2)=0.5, P(X=3)=0.3, find the distribution of Y = 2X + 1 and calculate E[Y].
Definition:

Transformed Random Variable: Y = g(X) where g is a function of X

Linearity of Expectation: E[aX + b] = aE[X] + b

Transformation Rule: P(Y = y) = P(X = x) where y = g(x)

Given
X: {P(1)=0.2, P(2)=0.5, P(3)=0.3}, Y = 2X + 1
Y values
{3, 5, 7}
E[Y]
E[2X + 1]
Step 1: Find Y values corresponding to X values

When X=1: Y = 2(1) + 1 = 3

When X=2: Y = 2(2) + 1 = 5

When X=3: Y = 2(3) + 1 = 7

Step 2: Determine probability distribution of Y

Since Y is uniquely determined by X:

P(Y=3) = P(X=1) = 0.2

P(Y=5) = P(X=2) = 0.5

P(Y=7) = P(X=3) = 0.3

Step 3: Calculate E[Y] using definition

E[Y] = 3×P(Y=3) + 5×P(Y=5) + 7×P(Y=7)

E[Y] = 3×0.2 + 5×0.5 + 7×0.3 = 0.6 + 2.5 + 2.1 = 5.2

Step 4: Verify using linearity of expectation

E[Y] = E[2X + 1] = 2E[X] + 1

E[X] = 1×0.2 + 2×0.5 + 3×0.3 = 0.2 + 1.0 + 0.9 = 2.1

E[Y] = 2(2.1) + 1 = 4.2 + 1 = 5.2 ✓

Y distribution: {3→0.2, 5→0.5, 7→0.3}, E[Y] = 5.2
Final answer:

The distribution of Y is P(Y=3)=0.2, P(Y=5)=0.5, P(Y=7)=0.3, and E[Y]=5.2

Applied rules:

• Transformation rule: P(Y=y) = P(X=x) where y=g(x)

• Linearity of expectation: E[aX + b] = aE[X] + b

• Probability preservation: Transformation doesn't change probability

Detailed Summary: Probability Distributions
\(E[X] = \sum x \cdot P(X=x) \text{ (discrete)}, E[X] = \int_{-\infty}^{\infty} x \cdot f(x) dx \text{ (continuous)}\)
Expected Value Formulas
Comprehensive definitions:

Random Variable (X): A function that maps outcomes of a random experiment to real numbers

Discrete Random Variable: Takes on a finite or countably infinite set of values

Continuous Random Variable: Takes on values in an interval or union of intervals

Probability Distribution: A specification of probabilities for all possible values of a random variable

Complete methodology:
  1. Problem identification: Determine if discrete or continuous, identify variable type
  2. Distribution selection: Choose appropriate distribution model
  3. Parameter estimation: Identify parameters from problem statement
  4. Probability calculation: Apply relevant formulas
Tip 1: Always check that probabilities sum to 1 (discrete) or PDF integrates to 1 (continuous).
Tip 2: For continuous distributions, P(X = a) = 0 for any specific value a.
Tip 3: Cumulative distribution function (CDF) is always non-decreasing and ranges from 0 to 1.
Tip 4: The expected value represents the long-term average value of the random variable.
Common distributions: Discrete: Binomial, Poisson, Geometric. Continuous: Normal, Uniform, Exponential.
Key properties: Expected value (mean), variance (spread), skewness (symmetry), kurtosis (tail behavior).
Essential formulas and rules:

• Discrete PMF: P(X = x) ≥ 0, Σ P(X = x) = 1

• Continuous PDF: f(x) ≥ 0, ∫f(x)dx = 1

• Expected value: E[X] = Σx·P(X=x) (discrete), E[X] = ∫xf(x)dx (continuous)

• Variance: Var(X) = E[X²] - (E[X])²

• Linearity of expectation: E[aX + b] = aE[X] + b

• Binomial: P(X = k) = C(n,k)·p^k·(1-p)^(n-k), E[X] = np, Var(X) = np(1-p)

Visualization: Probability Distribution Concepts
Distribution Characteristics
Compare the characteristics of discrete and continuous probability distributions:
Discrete: P(X = x) ≠ 0, PMF, Countable outcomes
Continuous: P(X = x) = 0, PDF, Uncountable outcomes
Both: E[X] = Expected value, Var(X) = Variance

Analysis: The chart shows probability distributions and their key characteristics.

  • Discrete distributions: Point probabilities, sum to 1, examples: dice, coin flips
  • Continuous distributions: Area probabilities, integrate to 1, examples: heights, time intervals
  • Both types have expected value and variance measures

Questions & Answers

Question: I don't understand why in continuous distributions, the probability of a specific value is zero. Can you explain this concept?

Answer: This is a fundamental concept that often confuses students. In continuous distributions:

  • There are infinitely many possible values within any range
  • Probability is represented by area under the PDF curve
  • A single point has no width, so the area under a single point is 0
  • Therefore, P(X = a) = 0 for any specific value a

Instead, we calculate probabilities for intervals: P(a ≤ X ≤ b) = ∫[a,b] f(x)dx

Think of it like trying to hit a specific point on a ruler with a dart - the chance of hitting exactly that point is essentially zero because there are infinitely many points on the ruler.

Question: When should I use binomial distribution versus other probability distributions?

Answer: Use binomial distribution when all of these conditions are met:

  • Fixed number of trials (n): You know in advance how many times the experiment will be repeated
  • Independent trials: The outcome of one trial doesn't affect others
  • Two possible outcomes: Success or failure (binary outcome)
  • Constant probability: The probability of success (p) remains the same for each trial

Examples: Coin flips, quality control in manufacturing, survey responses

If these conditions aren't met, consider other distributions like geometric (for first success), Poisson (for rare events), or normal (for continuous measurements).

Question: What's the difference between probability mass function (PMF) and probability density function (PDF)?

Answer: These are fundamental concepts that serve different purposes:

  • PMF (Discrete): Gives actual probability that X equals a specific value: P(X = x)
  • PDF (Continuous): Does NOT give probability directly; height of curve indicates relative likelihood
  • PMF: Probabilities are values on the function
  • PDF: Probabilities are areas under the curve

For discrete: P(X = a) = PMF(a) ≠ 0

For continuous: P(X = a) = 0, but P(a ≤ X ≤ b) = ∫[a,b] PDF(x)dx

Both must integrate/sum to 1 and be non-negative everywhere.

Question: How do transformations of random variables work? Why does E[aX + b] = aE[X] + b?

Answer: The linearity of expectation is a fundamental property that makes calculations much easier.

For discrete random variables:

E[aX + b] = Σ (ax + b) · P(X = x) = Σ [ax·P(X = x) + b·P(X = x)]

= a·Σ [x·P(X = x)] + b·Σ P(X = x) = a·E[X] + b·1 = aE[X] + b

For continuous random variables:

E[aX + b] = ∫ (ax + b) · f(x) dx = a∫ xf(x)dx + b∫ f(x)dx = aE[X] + b

This property holds regardless of the distribution of X, which is why it's so useful in probability theory.

Question: What is the cumulative distribution function (CDF) and how is it different from PMF/PDF?

Answer: The CDF is defined as F(x) = P(X ≤ x), giving the probability that the random variable takes a value less than or equal to x.

Key differences:

  • PMF/PDF: Give probability density at a specific point
  • CDF: Gives cumulative probability up to a point
  • PMF/PDF: Can be greater than 1 (especially for PDF with narrow peaks)
  • CDF: Always between 0 and 1
  • CDF: Always non-decreasing, approaches 0 as x → -∞, approaches 1 as x → ∞

For discrete: F(x) = Σ P(X = k) for k ≤ x

For continuous: F(x) = ∫[-∞,x] f(t)dt

Relationship: f(x) = dF(x)/dx (where derivative exists)