Calculate the Expected Number of Occurrences in Probability Experiments

Introduction

Key Concepts

1. Understanding Expected Value

The expected value, often denoted as $E(X)$, is a measure of the central tendency of a random variable. It represents the average outcome one would expect if an experiment were to be repeated numerous times under identical conditions. In the context of probability experiments, calculating the expected number of occurrences helps in predicting future events based on past data.

For a discrete random variable $X$ with possible values $x_1, x_2, ..., x_n$ occurring with probabilities $p_1, p_2, ..., p_n$ respectively, the expected value is calculated using the formula:

$$ E(X) = \sum_{i=1}^{n} x_i \cdot p_i $$

**Example:**

Consider rolling a fair six-sided die. Let $X$ be the random variable representing the number of dots rolled. The expected value is:

$$ E(X) = (1 \cdot \frac{1}{6}) + (2 \cdot \frac{1}{6}) + (3 \cdot \frac{1}{6}) + (4 \cdot \frac{1}{6}) + (5 \cdot \frac{1}{6}) + (6 \cdot \frac{1}{6}) = \frac{21}{6} = 3.5 $$

This means that, on average, one would expect to roll a 3.5 over many trials.

2. Calculating Expected Number of Occurrences

When dealing with multiple independent trials, the expected number of occurrences of an event can be calculated using the formula:

$$ E = n \cdot p $$

Where:

• $E$ is the expected number of occurrences.
• $n$ is the total number of trials.
• $p$ is the probability of the event occurring in a single trial.

**Example:**

If a coin is flipped 100 times, and the probability of getting heads in a single flip is 0.5, then the expected number of heads is:

$$ E = 100 \cdot 0.5 = 50 $$

3. Bernoulli Trials and Expected Value

Bernoulli trials are experiments where each trial has exactly two possible outcomes: success and failure. The expected number of successes in $n$ Bernoulli trials is given by:

$$ E = n \cdot p $$

Where:

• $n$ is the number of trials.
• $p$ is the probability of success on a single trial.

**Example:**

Suppose there is a 30% chance of rain on any given day. What is the expected number of rainy days in a 10-day period?

$$ E = 10 \cdot 0.3 = 3 $$

So, one would expect approximately 3 rainy days in 10 days.

4. Binomial Distribution

The binomial distribution models the number of successes in a fixed number of independent Bernoulli trials. The expected value of a binomially distributed random variable $X$ with parameters $n$ and $p$ is:

$$ E(X) = n \cdot p $$

This aligns with our earlier formula for the expected number of occurrences in Bernoulli trials.

5. Poisson Distribution and Expected Number of Occurrences

In scenarios where events occur continuously and independently at a constant average rate, the Poisson distribution is applicable. The expected number of occurrences in a Poisson distribution is represented by the parameter $\lambda$, which denotes the average number of events in the given interval.

**Formula:**

$$ E(X) = \lambda $$

**Example:**

If a call center receives an average of 5 calls per hour, the expected number of calls in one hour is 5.

6. Linearity of Expectation

The linearity of expectation states that the expected value of the sum of random variables is equal to the sum of their expected values, regardless of whether the variables are independent.

$$ E(X + Y) = E(X) + E(Y) $$

This principle allows for the calculation of expected values in more complex scenarios by breaking them down into simpler components.

7. Expected Value in Continuous Distributions

While the previous concepts dealt with discrete probability distributions, the expected number of occurrences in continuous distributions involves integration. For a continuous random variable $X$ with probability density function $f(x)$, the expected value is:

$$ E(X) = \int_{-\infty}^{\infty} x \cdot f(x) \, dx $$

However, in the context of the Cambridge IGCSE syllabus, focus is primarily on discrete distributions.

8. Application of Expected Number of Occurrences

Understanding the expected number of occurrences has practical applications in various fields such as quality control, finance, medicine, and engineering. For instance, in quality control, it helps in predicting the number of defective products in a production batch, thereby aiding in process improvements.

Practical Examples and Problems

**Example 1:**

A basketball player has a free-throw success rate of 80%. If the player shoots 50 free throws during practice, what is the expected number of successful shots?

$$ E = 50 \cdot 0.8 = 40 $$

So, the player is expected to make 40 successful free throws.

**Example 2:**

In a multiple-choice exam with 20 questions, each having 4 options, what is the expected number of correct answers if a student guesses all answers randomly?

$$ p = \frac{1}{4} \\ E = 20 \cdot \frac{1}{4} = 5 $$

Thus, the student is expected to answer 5 questions correctly by random guessing.

Advanced Concepts

1. Variance and Standard Deviation in Expected Value

Alongside the expected value, variance and standard deviation are essential measures of the spread of a probability distribution. While the expected value provides the central tendency, the variance quantifies the dispersion of outcomes around this central value.

The variance of a random variable $X$ is defined as:

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

For a binomial distribution, the variance is given by:

$$ Var(X) = n \cdot p \cdot (1 - p) $$

The standard deviation is the square root of the variance:

$$ \sigma = \sqrt{Var(X)} = \sqrt{n \cdot p \cdot (1 - p)} $$

**Example:**

Consider a scenario where a fair die is rolled 60 times. The probability of rolling a number greater than 4 (i.e., rolling a 5 or 6) is $p = \frac{2}{6} = \frac{1}{3}$. The expected number of such outcomes is:

$$ E(X) = 60 \cdot \frac{1}{3} = 20 $$

The variance is:

$$ Var(X) = 60 \cdot \frac{1}{3} \cdot \frac{2}{3} = 13.\overline{3} $$

The standard deviation is:

$$ \sigma = \sqrt{13.\overline{3}} \approx 3.65 $$

This indicates that the number of occurrences will typically vary by around 3.65 from the expected value.

2. Conditional Expectation

Conditional expectation involves calculating the expected value of a random variable given that another event has occurred. This concept is crucial when outcomes are dependent on specific conditions.

**Formula:**

$$ E(Y | X = x) = \sum_{y} y \cdot P(Y = y | X = x) $$

**Example:**

Suppose there are two bags of marbles. Bag A contains 4 red and 6 blue marbles, while Bag B contains 5 red and 5 blue marbles. A bag is chosen at random, and then a marble is drawn from the selected bag. What is the expected number of red marbles drawn?

Solution:

Probability of choosing Bag A: 0.5

Probability of choosing Bag B: 0.5

Expected number of red marbles:

$$ E(X) = 0.5 \cdot \frac{4}{10} + 0.5 \cdot \frac{5}{10} = 0.5 \cdot 0.4 + 0.5 \cdot 0.5 = 0.2 + 0.25 = 0.45 $$

So, the expected number of red marbles drawn is 0.45.

3. Expected Value of the Sum of Random Variables

When dealing with multiple random variables, the expected value of their sum is the sum of their expected values, provided that the variables are independent. This property is known as the linearity of expectation.

**Formula:**

$$ E(X + Y) = E(X) + E(Y) $$

**Example:**

Let $X$ be the number of successes in process A with $E(X) = 5$, and $Y$ be the number of successes in process B with $E(Y) = 3$. The expected number of total successes is:

$$ E(X + Y) = 5 + 3 = 8 $$

4. Applications in Real-World Scenarios

The calculation of expected number of occurrences extends beyond academic exercises and finds applications in various real-world scenarios:

• Quality Control: Determining the expected number of defective items in a production batch.
• Finance: Estimating expected returns on investments based on probabilistic models.
• Healthcare: Predicting the expected number of patients requiring a particular treatment.
• Risk Management: Assessing the expected number of claims in insurance portfolios.

5. Expected Number in Dependent Trials

While independence is a common assumption, real-life experiments may involve dependent trials. Calculating the expected number of occurrences in such cases requires understanding the nature and extent of the dependence between trials.

**Example:**

Consider drawing cards from a deck without replacement. The probability of drawing a specific card changes with each draw, making the trials dependent. Calculating the expected number requires adjusting the probability $p$ for each trial based on previous outcomes.

6. Multivariate Expected Values

In more complex experiments involving multiple variables, expected values can be extended to multivariate settings. This involves considering joint distributions and the interplay between various random variables.

**Example:**

Suppose two independent events occur with expected values $E(X) = 5$ and $E(Y) = 10$. The expected value of their product, assuming independence, is:

$$ E(XY) = E(X) \cdot E(Y) = 5 \cdot 10 = 50 $$

However, if $X$ and $Y$ are not independent, the calculation of $E(XY)$ becomes more involved.

7. Higher Moments and Skewness

Beyond the first moment (expected value), higher moments such as skewness measure the asymmetry of the probability distribution. Understanding these properties provides deeper insights into the behavior of random variables.

8. Expected Value in Mixed Distributions

In cases where a random variable follows different distributions under different conditions, calculating the expected value involves partitioning the scenarios and summing their contributions.

**Example:**

Assume a game where you roll a die. If you roll a 1-4, you gain \$10; if you roll a 5, you gain \$20; and if you roll a 6, you lose \$5. The expected gain is calculated as:

$$ E(X) = \left(\frac{4}{6} \cdot 10\right) + \left(\frac{1}{6} \cdot 20\right) + \left(\frac{1}{6} \cdot (-5)\right) = \frac{40}{6} + \frac{20}{6} - \frac{5}{6} = \frac{55}{6} \approx 9.17 $$

Thus, the expected gain per game is approximately \$9.17.

Mathematical Derivation and Proofs

Deriving expected values involves a solid understanding of probability theory. Below is a basic derivation for the expected value of a discrete random variable.

**Definition:**

For a discrete random variable $X$, the probability mass function (PMF) is defined as $P(X = x_i) = p_i$.

**Expected Value Formula:**

$$ E(X) = \sum_{i=1}^{n} x_i \cdot p_i $$

**Proof:** Start by considering the definition of expectation as the mean value of $X$ when the experiment is repeated a large number of times. Suppose the experiment is repeated $N$ times, and $n_i$ successes occur for outcome $x_i$. The relative frequency, $f_i$, approaches $p_i$ as $N$ becomes large: $$ f_i = \lim_{N \to \infty} \frac{n_i}{N} = p_i $$ The average value (sample mean) is: $$ \bar{x} = \frac{1}{N} \sum_{k=1}^{N} X_k $$ Where $X_k$ represents the outcome of the $k^{th}$ trial. Grouping outcomes by type: $$ \bar{x} = \frac{1}{N} \sum_{i=1}^{n} x_i \cdot n_i $$ Substituting $n_i = N \cdot p_i$: $$ \bar{x} = \frac{1}{N} \sum_{i=1}^{n} x_i \cdot (N \cdot p_i) = \sum_{i=1}^{n} x_i \cdot p_i = E(X) $$ Therefore, the expected value is as defined.

9. Wald's Equation

Wald's equation relates the expected value of the sum of a random number of random variables to the expected number of terms and the expected value of the individual terms. It is particularly useful in scenarios where the number of trials is itself a random variable.

**Formula:**

$$ E\left(\sum_{i=1}^{N} X_i\right) = E(N) \cdot E(X) $$

**Application:**

Suppose a photocopy machine breaks down randomly, and the number of copies made before a breakdown occurs is a random variable $N$ with $E(N) = 100$. If each copy made generates a profit of \$0.05, then the expected profit before a breakdown is:

$$ E(\text{Profit}) = E(N) \cdot E(X) = 100 \cdot 0.05 = \$5 $$

10. Non-integer Expected Values

The expected value does not need to be an integer, even if the random variable represents a count of occurrences. It represents a long-term average over many trials. Thus, in practice, the actual number of occurrences in any finite number of trials will be an integer, but the expected value can be a fractional number.

**Example:**

If the expected number of customers arriving at a store in an hour is 2.7, this implies that over many hours, the average number of customers per hour will approach 2.7, even though in any given hour, the actual number is an integer.

11. Expected Value in Game Theory

In game theory, expected values are used to determine the best strategies by evaluating the expected outcomes of different choices.

**Example:**

Consider a game where you can choose between two strategies: A and B. Strategy A results in a \$10 gain with probability 0.3 and a \$0 gain otherwise. Strategy B results in a \$5 gain with probability 0.6 and a \$0 gain otherwise.

Calculating expected values:

$$ E(A) = (0.3 \cdot 10) + (0.7 \cdot 0) = 3 $$ $$ E(B) = (0.6 \cdot 5) + (0.4 \cdot 0) = 3 $$

In this case, both strategies have the same expected gain, indicating that either could be chosen based on other criteria.

12. Expected Value under Different Probability Distributions

While the binomial and Poisson distributions are common, other distributions like the geometric and negative binomial also play roles in calculating expected occurrences in probability experiments.

Geometric Distribution: Models the number of trials until the first success. The expected number of trials is:

$$ E(X) = \frac{1}{p} $$

Negative Binomial Distribution: Models the number of trials until a specified number of successes occurs. The expected number of trials is:

$$ E(X) = \frac{r}{p} $$

**Example:**

For a fair coin ($p = 0.5$), the expected number of trials until the first head appears is:

$$ E(X) = \frac{1}{0.5} = 2 $$

13. Simulation of Expected Occurrences

Simulations can be used to empirically determine expected values by conducting numerous trials and calculating the average outcome. Software tools and programming languages like Python and R are commonly used for such simulations.

**Example with Simulation:**

To estimate the expected number of heads in 100 coin flips, perform multiple simulations where each simulation involves 100 flips, record the number of heads, and average the results over all simulations.

This empirical approach often corroborates the theoretical expected value derived using probability formulas.

14. Limitations of Expected Value

While expected value is a powerful concept, it has limitations. It does not account for the variability or risk associated with the occurrences. Two different distributions can have the same expected value but different variances, leading to different levels of uncertainty.

Additionally, in scenarios with infinite or undefined expected values, the concept may not provide meaningful insights.

15. Expected Value in Decision Making

In decision-making processes, expected values are used to evaluate options by quantifying the anticipated benefits or costs associated with each choice.

**Example:**

Suppose an investor can choose between two projects. Project X has an expected return of 8%, while Project Y has an expected return of 5%. Based solely on expected value, Project X is more attractive. However, if Project X also has a higher risk (variance) compared to Project Y, the investor must weigh expected returns against associated risks.

Comparison Table

Aspect	Binomial Distribution	Poisson Distribution
Number of Trials	Fixed number of trials ($n$)	Potentially infinite trials
Probability of Success	Constant probability ($p$) per trial	Average rate ($\lambda$)
Suitable For	Counting successes in a fixed number of independent trials	Counting occurrences in a fixed interval of time or space
Expected Value	$E(X) = n \cdot p$	$E(X) = \lambda$
Variance	$Var(X) = n \cdot p \cdot (1-p)$	$Var(X) = \lambda$

Summary and Key Takeaways