Question

Suppose we have two coins. One is fair, but the other one has two heads. We choose one of them at random and flip it. It comes up heads. (a) What is the probability the coin is fair? (b) Suppose we flip the same coin a second time. What is the probability that it comes up heads? (c) Suppose the coin comes up heads when flipped the second time. What is the probability the coin is fair?

Short answer

(a) 1/3; (b) 5/6; (c) 1/5

Step-by-step solution

Understand the Scenario

We have two coins: one fair coin (with a probability of 0.5 for heads) and one double-headed coin (with a probability of 1 for heads). A coin is randomly chosen and flipped, resulting in a head.

Define Events

Let F represent the event that the coin chosen is the fair coin, and D represent the event that it is the double-headed coin. Let H1 be the event of getting heads on the first flip, and H2 be the event of getting heads on the second flip.

Calculate Probability of Fair Coin Given Heads (Bayes' Theorem)

We need to find \( P(F \mid H_1) \). By Bayes' theorem: \( P(F \mid H_1) = \frac{P(H_1 \mid F) P(F)}{P(H_1 \mid F) P(F) + P(H_1 \mid D) P(D)} \). \( P(H_1 \mid F) = 0.5 \), \( P(F) = 0.5 \), \( P(H_1 \mid D) = 1 \), \( P(D) = 0.5 \). Calculate as follows: \( P(F \mid H_1) = \frac{0.5 \times 0.5}{0.5 \times 0.5 + 1 \times 0.5} = \frac{0.25}{0.25 + 0.5} = \frac{0.25}{0.75} = \frac{1}{3} \).

Probability of Heads on Second Flip

Given that the first flip was heads, determine the probability that the next flip is heads (P(H_2)). Use total probability: \( P(H_2 \mid H_1) = P(H_2 \mid F, H_1)P(F \mid H_1) + P(H_2 \mid D)P(D \mid H_1) \). \( P(H_2 \mid F, H_1) = 0.5 \), \( P(H_2 \mid D) = 1 \), \( P(D \mid H_1) = 1 - P(F \mid H_1) = \frac{2}{3} \). Calculate: \( P(H_2 \mid H_1) = 0.5 \times \frac{1}{3} + 1 \times \frac{2}{3} = \frac{1}{6} + \frac{2}{3} = \frac{1}{6} + \frac{4}{6} = \frac{5}{6} \).

Probability Coin is Fair Given Second Heads

We need \( P(F \mid H_2, H_1) \). Use Bayes' theorem: \( P(F \mid H_2, H_1) = \frac{P(H_2 \mid F, H_1) P(F \mid H_1)}{P(H_2 \mid H_1)} \). Calculate: \( P(F \mid H_2, H_1) = \frac{0.5 \times \frac{1}{3}}{\frac{5}{6}} = \frac{0.5/3}{5/6} = \frac{1/6}{5/6} = \frac{1}{5} \).

Key concepts

Probability

Probability is the mathematical way to measure how likely an event is to occur. In this context, when we talk about flipping a coin, we are concerned with the likelihood of landing on heads or tails.

- A fair coin, which has both a head and a tail, has a probability of 0.5 (50%) for getting heads and 0.5 for tails.

- A double-headed coin, however, has a probability of 1 for heads since it's always going to land heads up when flipped.

In any given random event, like flipping a coin, probabilities can assist in predicting outcomes. However, it’s crucial to remember that each flip is an independent event unless specifically stated otherwise, like in conditional probability scenarios.

Conditional Probability

Conditional probability is the likelihood of an event occurring, given that another event has already happened. It is very useful when dealing with complex scenarios where the outcome of one event affects another. In our problem with the coins, we use conditional probability to determine the likelihood that the chosen coin is fair or double-headed, given a particular outcome like heads showing up.

To calculate conditional probability, we often use Bayes' Theorem. This theorem helps us update our understanding of the probability of an event based on new evidence, thus refining our predictions:

• Bayes' Theorem Formula: \( P(A \,|\, B) = \frac{P(B \,|\, A) \cdot P(A)}{P(B)} \).
• This allows us to determine how probable an event A (the coin is fair) is, given event B (getting heads) has happened.

By applying this to the problem, for instance, if you initially believe the chance of picking the fair coin is 50%, and see heads, you can better assess what the remaining probability reflects.

Discrete Mathematics

Discrete mathematics is a branch of mathematics dealing with discrete elements that use algebra and arithmetic. It heavily focuses on distinct and separate values or objects, making it ideal for counting scenarios, such as probabilities in coin flips.

This field provides the basis for probability theory, including permutations, combinations, and events handling in scenarios similar to choosing a coin and calculating outcomes. It's the backbone for organizing and analyzing probability data. During coin-flip problems, discrete mathematics enables us to build sophisticated models to understand different statistical scenarios such as the one detailed in our exercise:
- Analyzing random variables to predict outcomes.
- Formulating expressions for probability using learned mathematical tools like Bayes' theorem. Discrete mathematics hence supplies the methods and principles for tackling and solving intricate probability problems and enhancing logical reasoning in decision-making processes.