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Three biased coins \(C_{1}, C_{2}, C_{3}\) lie on a table. Their respective probabilities of falling heads when tossed are \(\frac{1}{3}, \frac{2}{3}\), and 1 . A coin is picked at random, tossed, and observed to fall heads. Calculate the probability that it is \(C_{k}\) for each \(k=1,2,3\). Given that a coin has been tossed once and observed to fall heads, calculate the probability that a second throw of the same coin will also produce heads. The experiment is begun again with the same three coins. This time the coin selected is tossed twice and observed to fall heads both times. Calculate the probability that it is \(C_{k}\) for each \(k=1,2,3\). Given that a coin has been tossed twice and observed to fall heads both times, calculate the probability that a third throw of the same coin will also produce heads.

Step by step solution

01

Define Outcomes and Probabilities

We are dealing with three biased coins. Let's denote head probabilities as follows: Coin 1 with probability \( P(H_1) = \frac{1}{3} \), Coin 2 with \( P(H_2) = \frac{2}{3} \), and Coin 3 with \( P(H_3) = 1 \). Since we pick a coin at random initially, the probability of choosing any coin is \( \frac{1}{3} \).

02

Use Bayes' Theorem for Initial Head Outcome

Given a head on the first toss, we want the probability of selecting coin \( C_k \). According to Bayes' theorem, \( P(C_k \mid H) = \frac{P(H \mid C_k) \times P(C_k)}{P(H)} \). Here, \( P(H \mid C_1) = \frac{1}{3} \), \( P(H \mid C_2) = \frac{2}{3} \), and \( P(H \mid C_3) = 1 \). We calculate the total probability of observing heads, \( P(H) = \frac{1}{3}\times\frac{1}{3} + \frac{1}{3}\times\frac{2}{3} + \frac{1}{3}\times1 \).

03

Calculate Total Probability of Heads

Calculating \( P(H) \): \[P(H) = \frac{1}{9} + \frac{2}{9} + \frac{3}{9} = \frac{6}{9} = \frac{2}{3}\]

04

Probabilities for Each Coin After Head Outcome

Using Bayes' theorem for each coin: \[P(C_1 \mid H) = \frac{\frac{1}{9}}{\frac{2}{3}} = \frac{1}{6}\]\[P(C_2 \mid H) = \frac{\frac{2}{9}}{\frac{2}{3}} = \frac{1}{3}\]\[P(C_3 \mid H) = \frac{\frac{3}{9}}{\frac{2}{3}} = \frac{1}{2}\]

05

Probability of Heads on Second Toss

Given the probabilities for each coin after seeing the head, calculate the probability of heads on the second toss: \[P(H_2 \mid H) = \frac{1}{6} \times \frac{1}{3} + \frac{1}{3} \times \frac{2}{3} + \frac{1}{2} \times 1 \]\[P(H_2 \mid H) = \frac{1}{18} + \frac{2}{9} + \frac{1}{2} = \frac{1}{18} + \frac{4}{18} + \frac{9}{18} = \frac{14}{18} = \frac{7}{9}\]

06

Probability of Coin After Two Heads

Repeat Bayes' theorem if two heads are observed in a row. Calculate \( P(H \mid C_k)^2 \) for each coin: \[P(H \mid C_1)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \]\[P(H \mid C_2)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \]\[P(H \mid C_3)^2 = 1\]

07

Total Probability After Two Heads

Calculate the total probability of two heads: \[P(H \text{ twice}) = \frac{1}{3} \times \frac{1}{9} + \frac{1}{3} \times \frac{4}{9} + \frac{1}{3} \times 1 = \frac{1}{27} + \frac{4}{27} + \frac{9}{27} = \frac{14}{27}\]

08

Update Probabilities after Two Heads

Calculate the probability of each coin given two heads: \[P(C_1 \mid H \text{ twice}) = \frac{\frac{1}{27}}{\frac{14}{27}} = \frac{1}{14}\]\[P(C_2 \mid H \text{ twice}) = \frac{\frac{4}{27}}{\frac{14}{27}} = \frac{4}{14} = \frac{2}{7}\]\[P(C_3 \mid H \text{ twice}) = \frac{\frac{9}{27}}{\frac{14}{27}} = \frac{9}{14}\]

09

Probability of Heads on Third Toss

Finally, calculate heads on the third: \[P(H_3 \mid H \text{ twice}) = \frac{1}{14} \times \frac{1}{3} + \frac{2}{7} \times \frac{2}{3} + \frac{9}{14} \times 1 \]\[P(H_3 \mid H \text{ twice}) = \frac{1}{42} + \frac{4}{21} + \frac{9}{14} = \frac{1}{42} + \frac{8}{42} + \frac{27}{42} = \frac{36}{42} = \frac{6}{7}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Biased Coins

Biased coins are coins that do not have an equal chance of landing on heads or tails. Unlike fair coins, where the probability of getting heads or tails is 50% each, biased coins are skewed towards one side. In our exercise, we have three biased coins:

• Coin 1 with a probability of heads, \( \frac{1}{3} \).
• Coin 2 with a probability of heads, \( \frac{2}{3} \).
• Coin 3 that always lands on heads, probability of heads equals 1.

When dealing with biased coins, it is important to understand the probabilities associated with each coin separately. The bias influences the outcome of coin tosses, making it crucial to determine which coin we are using when trying to predict results.

Conditional Probability

Conditional probability is the likelihood of an event occurring given that another event has already occurred. It is an essential part of Bayes' theorem and helps us make inferences based on prior events. For example, in the exercise, we want to find the probability of a specific coin given heads are observed in a toss. We use Bayes' theorem to perform such calculations: \[ P(C_k \mid H) = \frac{P(H \mid C_k) \times P(C_k)}{P(H)} \] Where:

• \( P(C_k \mid H) \) is the probability of coin \( C_k \) given a head outcome.
• \( P(H \mid C_k) \) is the probability of heads given we picked coin \( C_k \).
• \( P(C_k) \) is the prior probability of picking coin \( C_k \).
• \( P(H) \) is the overall probability of observing heads.

Using this setup, we can find insights into which coin is more likely when certain outcomes are observed.

Probability of Multiple Events

The probability of multiple events refers to calculating the likelihood of two or more events happening in succession. In tasks involving coin tossing, it's important to consider the sequence of outcomes. For our biased coins problem, this is evident when assessing probabilities over multiple tosses. For instance, the probability of getting heads twice with one coin, given heads the first time, employs repetition of conditional probabilities as calculated via Bayes' theorem. First, we calculate the probability of the first outcome, then build upon it for subsequent outcomes. Here's an example calculation approach:

• Determine the probability of heads twice: \( P(H \text{ twice}) \)
• Factor in each coin's head sequence probability squared: \( P(H \mid C_k)^2 \)
• Apply results to calculate new probabilities per coin: \( P(C_k \mid H \text{ twice}) \)

Such calculations allow us to see how outcomes evolve with each event and are crucial in scenarios with multiple biased coins.