Question

(a) A weighted coin has probability of \(\frac{2}{3}\) of showing heads and \(\frac{1}{3}\) of showing tails. Find the probabilities of \(h h, h t, t h\) and \(t t\) in two tosses of the coin. Set up the sample space and the associated probabilities. Do the probabilities add to 1 as they should? What is the probability of at least one head? What is the probability of two heads if you know there was at least one head? (b) For the coin in (a), set up the sample space for three tosses, find the associated probabilities, and use it to answer the questions in Problem 2.12 .

Short answer

Probabilities for two tosses: \( P(HH) = \frac{4}{9}, P(HT) = \frac{2}{9}, P(TH) = \frac{2}{9}, P(TT) = \frac{1}{9} \. Probability of at least one head: \( \frac{8}{9} \. Probability of two heads given at least one head: \( \frac{1}{2} \.

Step-by-step solution

Understand the Problem

A biased coin has a probability \( \frac{2}{3} \) of showing heads (H) and \( \frac{1}{3} \) of showing tails (T). The task is to find probabilities for each possible outcome in two and three tosses, and answer specific questions about probabilities.

Set up the sample space for two tosses

In two tosses, the sample space contains: \( \{ HH, HT, TH, TT \} \).

Calculate probabilities for each outcome in two tosses

Calculate the probability for each outcome:\( P(HH) = \frac{2}{3} \times \frac{2}{3} = \frac{4}{9} \)\( P(HT) = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9} \)\( P(TH) = \frac{1}{3} \times \frac{2}{3} = \frac{2}{9} \)\( P(TT) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} \)

Verify probabilities sum to 1

Sum up all probabilities: \( \frac{4}{9} + \frac{2}{9} + \frac{2}{9} + \frac{1}{9} = 1 \)

Calculate probability of at least one head in two tosses

Use complementary probability: \( P(at\text{ least one } H) = 1 - P(TT) = 1 - \frac{1}{9} = \frac{8}{9} \)

Calculate probability of two heads given at least one head

Conditional probability: \( P(HH \mid at\text{ least one } H) = \frac{P(HH)}{P(at\text{ least one } H)} = \frac{\frac{4}{9}}{\frac{8}{9}} = \frac{1}{2} \)

Set up the sample space for three tosses

In three tosses, the sample space contains: \( \{ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT \} \)

Calculate probabilities for each outcome in three tosses

Calculate the probability for each outcome: \( P(HHH) = \left( \frac{2}{3} \right)^3 = \frac{8}{27} \) \( P(HHT) = P(HTH) = P(THH) = \frac{2}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{4}{27} \) each \( P(HTT) = P(THT) = P(TTH) = \frac{2}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{2}{27} \) each \( P(TTT) = \left( \frac{1}{3} \right)^3 = \frac{1}{27} \)

Verify probabilities sum to 1

Sum up all probabilities: \( \frac{8}{27} + 3 \times \frac{4}{27} + 3 \times \frac{2}{27} + \frac{1}{27} = 1 \)

Key concepts

Sample Space

In probability theory, a sample space is the set of all possible outcomes of a particular experiment. In simpler terms, it's a list of everything that can happen when you perform an action, such as flipping a coin or rolling a die. For example, if you toss a biased coin twice, the sample space would be: \( \{HH, HT, TH, TT\} \). Each pair in the set represents a possible outcome from the two coin tosses. For three tosses, the sample space expands to include all combinations of heads and tails across the three events: \( \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\} \). Essentially, the sample space helps us understand all potential results and lays the groundwork for calculating probabilities.

Conditional Probability

Conditional probability is the likelihood of an event occurring given that another event has already happened. It helps in updating the probability of an event based on new information. Let's consider the step-by-step solution provided earlier: if you know there was at least one head in two coin tosses, you can find the probability of getting two heads. This is calculated using the formula: \( P(A | B) = \frac{P(A \cap B)}{P(B)} \). Here, \( P(HH | at\text{ least one } H) = \frac{P(HH)}{P(at\text{ least one } H)} = \frac{\frac{4}{9}}{\frac{8}{9}} = \frac{1}{2} \). Conditional probability is crucial for making informed predictions based on given conditions.

Complementary Probability

Complementary probability is used to find the likelihood of an event not happening by subtracting the probability of the event happening from 1. This is often easier and quicker than calculating the probability directly. For instance, in the exercise solution, the probability of getting at least one head in two coin tosses is explored. To find this, we calculate the complementary probability of getting no heads (both tails), which is \( P(TT) = \frac{1}{9} \). Then, the complementary probability is: \( P(at\text{ least one } H) = 1 - P(TT) = 1 - \frac{1}{9} = \frac{8}{9} \). Complementary probability simplifies complex probability calculations.

Biased Coin

A biased coin is one that doesn't have an equal chance of landing on heads or tails when flipped. Instead, it has a probability distribution that favors one outcome over another. For example, in the given exercise, the biased coin has a probability \( \frac{2}{3} \) of showing heads and \( \frac{1}{3} \) of showing tails. This weighted probability influences the outcomes and their respective probabilities. In the sample space of two tosses, the different combinations (HH, HT, TH, and TT) all have probabilities derived from the biased chances of heads and tails. Understanding a biased coin helps in accurately calculating probabilities when the events are not equally likely.