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Chapter 3: Problem 43

There are 3 coins in a box. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of the time. When one of the 3 coins is selected at random and flipped, it shows heads. What is the probability that it was the two-headed coin?

Short Answer

Expert verified
The probability that the two-headed coin was selected, given that a head was flipped, is approximately 0.444 or 44.4%.

Step by step solution

01

Understand Bayes' Theorem

Bayes' Theorem states that for events A and B: P(A|B) = \(\frac{P(B|A) * P(A)}{P(B)}\) In our case, event A is selecting the two-headed coin, and event B is flipping a head. We want to find P(A|B), which is the probability of selecting the two-headed coin given that a head was flipped.
02

Calculate the probabilities

Let's calculate the probabilities we need for Bayes' Theorem: 1. P(A): The probability of selecting the two-headed coin is 1/3, since there are three coins and each has an equal chance of being selected. 2. P(B|A): If we have selected the two-headed coin (event A), the probability of flipping a head (event B) is 100%, or 1. 3. P(B): The probability of flipping a head, considering all the coins: P(B) = P(B|A) * P(A) + P(B|A') * P(A') + P(B|A'') * P(A'') where A', A'' are the other two coins (fair coin and biased coin) We already have the probability of flipping a head given that the two-headed coin was selected, P(B|A) = 1. Now let's find the probabilities for the other coins: - Fair coin (A'): P(B|A') = 0.5 (since it's a fair coin) and P(A') = 1/3 (equal chance of being selected). - Biased coin (A''): P(B|A'') = 0.75 (heads 75% of the time) and P(A'') = 1/3 (equal chance of being selected). Now, we can calculate the overall probability of flipping a head: P(B) = P(B|A) * P(A) + P(B|A') * P(A') + P(B|A'') * P(A'') = 1 * (1/3) + 0.5 * (1/3) + 0.75 * (1/3)
03

Apply Bayes' Theorem

Now that we have all the probabilities, we can use Bayes' Theorem to find the probability of selecting the two-headed coin given that a head was flipped: P(A|B) = \(\frac{P(B|A) * P(A)}{P(B)}\) P(A|B) = \(\frac{1 * (1/3)}{1*(1/3) + 0.5*(1/3) + 0.75*(1/3)}\)
04

Simplify the expression

Simplifying the expression from Step 3, we will find the probability that the two-headed coin was selected: P(A|B) = \(\frac{1*(1/3)}{1*(1/3) + 0.5*(1/3) + 0.75*(1/3)}\) = \(\frac{1}{1 + 0.5 + 0.75}\) = \(\frac{1}{2.25}\) P(A|B) 鈮� 0.444 So, the probability that the two-headed coin was selected, given that a head was flipped, is approximately 0.444 or 44.4%.

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Key Concepts

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

Conditional Probability
At its core, conditional probability is about understanding the likelihood of an event occurring given that another event has already happened. Think of it like updating your beliefs with new evidence. When flipping a coin, for instance, the probability of it being heads or tails is initially 50-50. But, if you knew the coin was two-headed, your prediction would change dramatically, right? That鈥檚 conditional probability in action.

In the context of the given exercise, we're interested in the probability of having chosen the two-headed coin (\( P(A) \)), provided we've already seen a head (\( P(B) \)). This scenario is a perfect illustration of conditional probability because the outcome of the flipped coin influences our belief about which coin was chosen.
Probability Theory
Now, let's delve into probability theory, which forms the bedrock of statistical analysis and helps us predict the likelihood of various outcomes. It's the mathematical framework for quantifying uncertainty. Probability theory spans from simple events like tossing coins or rolling dice to complex random events in finance and science.

To compute the desired probability in our exercise, we apply the concepts of probability theory, using Bayes' Theorem to intertwine the odds of random draws (choosing a coin) with conditional outcomes (the coin showing heads). This blend of probability theory and real-world scenarios exemplifies the immense value and applicability of understanding such mathematical concepts.
Random Events
And finally, we address the concept of random events, which are situations where the outcome is not certain and can vary in an unpredictable manner. Each coin toss is independent, and the outcome (heads or tails) is random, albeit with known probabilities. In our exercise, selecting one of the three coins is a random event too.

However, once we know that a head has come up, it's no longer a process of random selection among all coins; it's now a weighted consideration where our two-headed coin has become a more likely culprit. This demonstrates how random events can change their nature when new information is added to the equation - a key idea within probability theory.

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Most popular questions from this chapter

Suppose that \(E\) and \(F\) are mutually exclusive events of an experiment. Show that if independent trials of this experiment are performed, then \(E\) will occur before \(F\) with probability \(P(E) /[P(E)+P(F)]\).

3.25. The following method was proposed to estimate the number of people over the age of 50 who reside in a town of known population 100,000: "As you walk along the streets, keep a running count of the percentage of people you encounter who are over \(50 .\) Do this for a few days; then multiply the percentage you obtain by 100,000 to obtain the estimate." Comment on this method. Hint: Let \(p\) denote the proportion of people in the town who are over \(50 .\) Furthermore, let \(\alpha_{1}\) denote the proportion of time that a person under the age of 50 spends in the streets, and let \(\alpha_{2}\) be the corresponding value for those over \(50 .\) What quantity does the method suggested estimate? When is the estimate approximately equal to \(p ?\)

Suppose we have 10 coins such that if the \(i\) th coin is flipped, heads will appear with probability \(i / 10, i=\) \(1,2, \ldots, 10 .\) When one of the coins is randomly selected and flipped, it shows heads. What is the conditional probability that it was the fifth coin?

Die \(A\) has 4 red and 2 white faces, whereas die \(B\) has 2 red and 4 white faces. A fair coin is flipped once. If it lands on heads, the game continues with die \(A\); if it lands on tails, then die \(B\) is to be used. (a) Show that the probability of red at any throw is \(\frac{1}{2}\) (b) If the first two throws result in red, what is the probability of red at the third throw? (c) If red turns up at the first two throws, what is the probability that it is die \(A\) that is being used?

Suppose that we want to generate the outcome of the flip of a fair coin, but that all we have at our disposal is a biased coin that lands on heads with some unknown probability \(p\) that need not be equal to \(\frac{1}{2} .\) Consider the following procedure for accomplishing our task: 1\. Flip the coin. 2\. Flip the coin again. 3\. If both flips land on heads or both land on tails, return to step 1. 4\. Let the result of the last flip be the result of the experiment. (a) Show that the result is equally likely to be either heads or tails. (b) Could we use a simpler procedure that continues to flip the coin until the last two flips are different and then lets the result be the outcome of the final flip?
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