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Chapter 5: Problem 7

Suppose that coin I has probability \(0.7\) of coming up heads, and coin 2 has probability \(0.6\) of coming up heads. If the coin flipped today comes up heads, then we select coin 1 to flip tomorrow, and if it comes up tails, then we select coin 2 to flip tomorrow. If the coin initially flipped is equally likely to be coin 1 or coin 2, then what is the probability that the coin flipped on the third day after the initial flip is coin \(1 ?\)

Short Answer

Expert verified
The probability that the coin flipped on the third day after the initial flip is coin 1 is \(0.665\).

Step by step solution

01

Draw and analyze the probability tree for flipping the coins

Draw a probability tree showing all possible paths to the third day's flip. Starting from the initial flip, there are two possible outcomes: Heads (H) or Tails (T). Depending on the outcome, we select either coin 1 or coin 2 for the next day's flip.
02

Calculate the probability of each path in the tree

For each path, multiply the probabilities of each step on that path. There are 8 possible paths to reach the third day's flip, and their respective probabilities are: 1. Coin 1 -> Coin 1 -> Coin 1 (H, H): \(\frac{1}{2} * 0.7 * 0.7 = 0.245\) 2. Coin 1 -> Coin 1 -> Coin 2 (H, T): \(\frac{1}{2} * 0.7 * 0.3 = 0.105\) 3. Coin 1 -> Coin 2 -> Coin 1 (T, H): \(\frac{1}{2} * 0.3 * 0.6 = 0.09\) 4. Coin 1 -> Coin 2 -> Coin 2 (T, T): \(\frac{1}{2} * 0.3 * 0.4 = 0.06\) 5. Coin 2 -> Coin 1 -> Coin 1 (H, H): \(\frac{1}{2} * 0.6 * 0.7 = 0.21\) 6. Coin 2 -> Coin 1 -> Coin 2 (H, T): \(\frac{1}{2} * 0.6 * 0.3 = 0.09\) 7. Coin 2 -> Coin 2 -> Coin 1 (T, H): \(\frac{1}{2} * 0.4 * 0.6 = 0.12\) 8. Coin 2 -> Coin 2 -> Coin 2 (T, T): \(\frac{1}{2} * 0.4 * 0.4 = 0.08\)
03

Calculate the probability that the coin flipped on the third day is coin 1

Add the probabilities of all paths that end up with coin 1 on the third day: Prob(Third day's flip is Coin 1) = Prob(Path 1) + Prob(Path 3) + Prob(Path 5) + Prob(Path 7) = 0.245 + 0.09 + 0.21 + 0.12 = 0.665 The probability that the coin flipped on the third day after the initial flip is coin 1 is 0.665.

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Key Concepts

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

Probability Tree
A probability tree is a visual tool used to map out all possible outcomes of a sequence of events and their corresponding probabilities. It's particularly useful for understanding complex scenarios where events are linked and outcomes depend on previous results.

Let's apply this to our coin flipping exercise. Imagine a branching diagram starting with the initial flip. Each branch represents each outcome: heads or tails. Each subsequent flip then branches off from the preceding one, allowing us to trace all possible scenarios to the third day's flip.

By calculating the probability of each path by multiplying the probabilities of each step and adding them together, we can determine the likelihood of any given outcome. It's this organized, step-wise approach that can transform a tangled possibility into an easy-to-follow sequence of scenarios, where we can quickly pinpoint the probability that the coin flipped on the third day is coin 1.
Independent Events
When dealing with probabilities, it's vital to identify if the events are independent - the outcome of one event does not affect the outcome of another. In the coin flip exercise, whether we flip coin 1 or 2 each day is influenced by the previous day's result, technically making them dependent. However, for a true independent event, each flip would have no bearing on the next.

Understanding this concept of independence is crucial because it determines how we calculate combined probabilities. When events are independent, the probability of both occurring is the product of their individual probabilities. Exploring this concept alongside probability trees allows students to distinguish between dependent and independent scenarios, fostering deeper comprehension of the complexities within probability theory.
Bayesian Probability
Bayesian probability is a level up in our understanding of probabilities. It involves updating the probability of a hypothesis as more evidence or information becomes available.

In the context of our coin flip problem, imagine we didn't know the likelihood of each coin being flipped initially and only observed the outcomes. Using Bayesian techniques, we could update our beliefs about which coin it was with each successive flip. This form of probability is dynamic, representing a shift from static, initial beliefs to informed updates that reflect new data. It is a powerful framework for decision-making in uncertain conditions across various fields from science to economics.

While the exercise given already provides probabilities, a Bayesian approach would be ready to revise those probabilities as additional flips reveal more information about the coins' behavior.

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Most popular questions from this chapter

A particle moves among \(n+1\) vertices that are situated on a circle in the following manner: At cach step it moves one step either in the clockwise direction with probability \(p\) or the counterclockwise direetion with probability \(q=1-p\). Starting at a specified state, call it state 0 , let \(T\) be the time of the first return to state 0 . Find the probability that all states have been visited by time \(T\).

For a series of dependent trials the probability of success on any trial is \((k+1) /(k+2)\) where \(k\) is equal to the number of successes on the previous two trials. Compute lim \(_{n \rightarrow \infty} P\) (success on the \(n\) th trial).

16\. Let \(Y_{n}\) be the sum of \(n\) independent rolls of a fair dic. Find $$ \lim _{n \rightarrow \infty} P\left[Y_{n} \text { is a multiple of } 13\right] $$ Define an appropriate Markov chain and apply the results of Exercise 14.

Consider a branching process having \(\mu<1 .\) Show that if \(X_{0}=1\), then the expected number of individuals that ever exist in this population is given by \(1 /(1-\mu) .\) What if \(X_{0}=n ?\)

Three white and three black balls are distributed in two urns in such a way that each contains three balls. We say that the system is in state \(l\), \(i=0,1,2,3\), if the first urn contains \(i\) white balls. At cach step, we draw one ball from each urn and place the ball drawn from the first urn into the second, and conversely with the ball from the second urn. Let \(X_{n}\) denote the state of the system after the \(n\) th step. Explain why \(\left[X_{n}, n=0,1,2, \ldots\right]\) is a Markov chain and calculate its transition probability matrix.
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