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Coin 1 comes up heads with probability \(0.6\) and coin 2 with probability 0.5. A coin is continually flipped until it comes up tails, at which time that coin is put aside and we start flipping the other one. (a) What proportion of flips use coin \(1 ?\) (b) If we start the process with coin 1 what is the probability that \(\operatorname{coin} 2\) is used on the fifth flip?

Short Answer

Expert verified
The proportion of flips using coin 1 is approximately \(0.5714\), and the probability that coin 2 is used on the fifth flip, given that we start with coin 1, is approximately \(0.0864\).

Step by step solution

01

Part (a): Proportion of flips using coin 1

To find out the proportion of flips using coin 1, we must first calculate the probability of switching to the other coin in both cases. For coin 1, the probability of getting tails and switching is \((1-0.6)=0.4.\) For coin 2, the probability of getting tails and switching is \((1-0.5)=0.5.\) Since these probabilities are independent of each other, we can use geometric probabilities to find out the proportion of flips using coin 1. Let \(p_1\) be the probability of flipping coin 1, then we have \(p_1 = 0.4 + 0.6\times 0.5\times p_1\) Now we can solve for \(p_1\).
02

Solve for the proportion of flips using coin 1

From the equation above, we have: \(p_1(1 - 0.6\times 0.5) = 0.4\) Now, let's solve for \(p_1\): \(p_1 = \dfrac{0.4}{1 - 0.6 \times 0.5} = \dfrac{0.4}{1 - 0.3} = \dfrac{0.4}{0.7}\) So the proportion of flips using coin 1 is approximately \(0.5714.\)
03

Part (b): Find the probability that coin 2 is used on the fifth flip

To find the probability that coin 2 is used on the fifth flip, we need to calculate the probability that coin 1 is flipped four times before switching to coin 2 for the fifth flip. For this to happen, coin 1 must come up heads three times (probability = \((0.6)^3\)) and then tails (probability = \(0.4\)) to switch to coin 2.
04

Calculate the probability that coin 2 is used on the fifth flip

Following the scenario conditions described above, we can find the probability that coin 2 is used on the fifth flip: \(\mathrm{Probability} = (0.6)^3\times 0.4\) Now, we perform the multiplication: \(\mathrm{Probability} = (0.216) \times 0.4 = 0.0864\) So, starting with coin 1, the probability that coin 2 is used on the fifth flip is approximately \(0.0864\).

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Key Concepts

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

Probability Models
When we talk about probability models, we're looking at frameworks that allow us to calculate the likelihood of various outcomes. They're essentially mathematical representations of real-world random processes. A classic example, which relates to our problem here, is flipping coins. In probability models, we consider each possible outcome (like getting a head or a tail) and assign probabilities to them based on certain conditions.

There are different types of probability models, but in the context of our problem, we work with a simple model where each coin flip is an independent event with its own probability. For example, coin 1 coming up heads has a probability of 0.6, while coin 2 has a different probability of 0.5. By using these individual probabilities, we can build a model to predict how often we'll flip each coin and the chances of switching from one coin to another during a series of flips.
Independent Events
Independent events are fundamental to understanding probability theory. Two events are independent if the outcome of one does not affect the outcome of the other. In our exercise, when we flip a coin, it doesn't matter what happened in previous flips; each flip's result is independent of the others.

Knowing this, we can calculate probabilities of sequences of flips by simply multiplying the individual probabilities together. For instance, the probability that coin 1 shows heads three times in a row is calculated by multiplying the probability of getting heads on a single flip, 0.6, by itself three times: \(0.6\times0.6\times0.6\). This is possible precisely because each flip is an independent event.
Geometric Distribution
The geometric distribution describes the number of trials it takes for a success to occur in a sequence of independent and identically distributed Bernoulli trials. In layman's terms, it tells us how many times we need to flip a coin before seeing the outcome we're looking for, like flipping tails for the first time. The distribution is 'geometric' because the probabilities decrease geometrically.

In the context of our example, if we're waiting for a tail to switch from coin 1 to coin 2, we're dealing with a geometric distribution scenario. Each flip brings us a certain chance of 'success' (switching the coin), and the geometric distribution helps us define the probability of that success happening on the first flip, the second flip, and so on. The geometric distribution applies to our part (b), where we're calculating the probability that coin 2 is used on the fifth flip. We determine this by looking at the probability of getting three heads in a row when flipping coin 1 (success for coin 1) followed by a tail (which leads to the success of switching to coin 2).

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Most popular questions from this chapter

Consider a Markov chain with states \(0,1,2,3,4\). Suppose \(P_{0,4}=1\); and suppose that when the chain is in state \(l, i>0\), the next state is equally likely to be any of the states \(0,1, \ldots, i-1\). Find the limiting probabilities of this Markov chain.

Suppose that a population consists of a fixed number, say, \(m\), of genes in any generation. Each gene is one of two possible genetic types. If any generation has exactly \(i\) (of its \(m\) genes being type 1 , then the next generation will have \(j\) type 1 (and \(m-j\) type 2) genes with probability $$ \left(\begin{array}{c} m \\ j \end{array}\right)\left(\frac{i}{m}\right)^{\prime}\left(\frac{m-i}{m}\right)^{m-j}, \quad j=0,1, \ldots, m $$ Let \(X_{n}\) denote the number of type 1 genes in the \(n\) th generation, and assume that \(X_{0}=i .\) (a) Find \(E\left[X_{n}\right]\). (b) What is the probability that eventually all the genes will be type \(1 ?\)

\(M\) balls are initially distributed among \(m\) urns. At each stage one of the balls is selected at random, taken from whichever urn it is in, and then placed, at random, in one of the other \(M-1\) urns. Consider the Markov chain whose state at any time is the vector \(\left(n_{1}, \ldots, n_{m}\right)\) where \(n_{i}\) denotes the number of balls in urn \(1 .\) Guess at the limiting probabilities for this Markov chain and then verify your guess and show at the same time that the Markov chain is time reversible.

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 \(1 ?\)

(a) Show that the limiting probabilities of the reversed Markov chain are the same as for the forward chain by showing that they satisfy the equations $$ \pi_{j}=\sum_{i} \pi_{i} Q_{u} $$ (b) Give an intuitive explanation for the result of part (a).

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