Question

Show that the expected number of heads in a single toss of a coin is \(\frac{1}{2}\). Show in two ways that the expected number of heads in two tosses of a coin is 1: (a) Let \(x=\) number of heads in two tosses and find \(\bar{x}\). (b) Let \(x=\) number of heads in toss 1 and \(y=\) number of heads in toss 2 ; find the average of \(x+y\) by Problem \(9 .\) Use this method to show that the expected number of heads in \(n\) tosses of a coin is \(\frac{1}{2} n\).

Short answer

The expected number of heads in two tosses is 1, and in n tosses it is \(\frac{1}{2} n\).

Step-by-step solution

Expected number of heads in a single toss

For a single coin toss, the outcomes are heads (H) and tails (T). The probability of heads is \(\frac{1}{2}\) and the probability of tails is \(\frac{1}{2}\). The expected value (E) is given by \(\text{E}(\text{X}) = \text{Sum of} (\text{outcome} \times \text{probability}) \ = 1 \times \frac{1}{2} + 0 \times \frac{1}{2} = \frac{1}{2} \).

Expected number of heads in two tosses (Part a)

When a coin is tossed twice, the possible outcomes are HH, HT, TH, TT. We can represent the number of heads as X, which can take values 0 (TT), 1 (HT or TH), or 2 (HH). The probabilities are: P(X=0) = \(\frac{1}{4}\), P(X=1) = \(\frac{2}{4} = \frac{1}{2}\), P(X=2) = \(\frac{1}{4}\). Now, we calculate the expected value \(\text{E}(X) = 0 \times \frac{1}{4} + 1 \times \frac{1}{2} + 2 \times \frac{1}{4} = 0 + \frac{1}{2} + \frac{1}{2} = 1\).

Expected number of heads in two tosses (Part b)

Let X represent the number of heads in the first toss and Y represent the number of heads in the second toss. For both tosses: \(\text{E}(X) = \text{E}(Y) = \frac{1}{2} \). By the linearity property of expectation: \(\text{E}(X + Y) = \text{E}(X) + \text{E}(Y) = \frac{1}{2} + \frac{1}{2} = 1\).

Expected number of heads in n tosses

Using the same method, let X_1, X_2, ..., X_n be the number of heads in each of the n tosses. The expected value for each toss is \(\text{E}(X_i) = \frac{1}{2}\). By the linearity of expectation: \(\text{E}(X_1 + X_2 + ... + X_n) = \text{E}(X_1) + \text{E}(X_2) + ... + \text{E}(X_n) = n \times \frac{1}{2} = \frac{1}{2} n\).

Key concepts

Probability Theory

Probability theory is the branch of mathematics that deals with the likelihood of different events occurring. It's used to predict the outcomes of random events and is essential in fields like statistics, finance, and science. One key aspect of probability theory is understanding the concept of probability itself, which measures how likely an event is to happen.
For example, in a coin toss, there are two possible outcomes: heads (H) or tails (T). Each outcome has an equal chance of occurring, so the probability of getting heads is \(\frac{1}{2}\), and the probability of tails is also \(\frac{1}{2}\). These probabilities must add up to 1, as they cover all possible outcomes.
Probability theory uses formulas and models to predict these outcomes over numerous trials, giving us insights into long-term trends.

Expected Value Calculation

The expected value is a fundamental concept in probability theory. It represents the average that you would expect to get if you could repeat an experiment infinitely many times. It's calculated as the sum of all possible outcomes, each multiplied by their respective probabilities.
For instance, for a single coin toss, the expected value of getting heads can be computed as: \[ \text{E}(\text{X}) = (1 \times \frac{1}{2}) + (0 \times \frac{1}{2}) = \frac{1}{2} \] Here, we consider getting heads as 1, and tails as 0.
This method can be extended to more complex scenarios, like calculating the expected number of heads in multiple coin tosses. Understanding how to calculate the expected value helps in predicting expected outcomes in various probabilistic situations.

Coin Toss Experiment

The coin toss experiment is a classic example used to explain probability concepts. It involves tossing a coin to see whether it lands on heads or tails. Each toss is an independent event, meaning the result of one toss does not affect the next.
In this exercise, we analyze the expected number of heads in one, two, and multiple coin tosses. With one toss, the probability of getting heads is \( \frac{1}{2} \), so the expected value is \(\frac{1}{2} \). For two tosses, there are four possible outcomes: HH, HT, TH, and TT. The expected number of heads can be calculated using these outcomes and their probabilities: \[ \text{E}(X) = (0 \times \frac{1}{4}) + (1 \times \frac{1}{2}) + (2 \times \frac{1}{4}) = 0 + \frac{1}{2} + \frac{1}{2} = 1 \]
By conducting this experiment, we can better understand the mechanics of probability and expected value in a straightforward scenario.

Linearity of Expectation

Linearity of expectation is a useful property in probability theory that simplifies calculations of expected values for complex scenarios. It states that the expected value of the sum of random variables is equal to the sum of their individual expected values.
In other words, if you have random variables \(X_1, X_2, ..., X_n\), then \[ \text{E}(X_1 + X_2 + ... + X_n) = \text{E}(X_1) + \text{E}(X_2) + ... + \text{E}(X_n) \] This property is particularly handy in the coin toss experiment. For two tosses, let \(X\) represent heads in the first toss and \(Y\) represent heads in the second toss. Each has an expected value of \( \frac{1}{2} \). Therefore, using the linearity of expectation: \[ \text{E}(X + Y) = \text{E}(X) + \text{E}(Y) = \frac{1}{2} + \frac{1}{2} = 1 \]
This method extends to \(n\) tosses, where the expected number of heads is \( \frac{1}{2} n \). Linearity makes computing the expected values manageable, even in larger and more complex probabilistic experiments.