/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 A coin, which lands on heads wit... [FREE SOLUTION] | 91Ó°ÊÓ

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A coin, which lands on heads with probability \(p\), is continually flipped. Compute the expected number of flips that are made until a string of \(r\) heads in a row is obtained. HINT: Condition on the time of the first occurrence of tails, to obtain the equation $$ E[X]=(1-p) \sum_{i=1}^{r} p^{i-1}(i+E[X])+(1-p) \sum_{i=r+1}^{x} p^{i-1} r $$ Simplify and solve for \(E[X]\).

Short Answer

Expert verified
The expected number of flips to obtain \(r\) heads in a row for a biased coin with probability \(p\) of heads is given by the formula: \(E[X]=1 + rp^{r}\).

Step by step solution

01

1. Understand the provided equation

The provided equation is a conditional expected value equation: $$ E[X]=(1-p) \sum_{i=1}^{r} p^{i-1}(i+E[X])+(1-p) \sum_{i=r+1}^{x} p^{i-1} r $$ The first term of this equation represents the probability of landing tails (1-p) multiplied by the situation when the first tails appear within the first r flips. The second term represents the probability of landing tails (1-p) multiplied by the situation when the first tails appear after r flips.
02

2. Modify the summation limits of the equation to simplify

In order to simplify the given equation, we need to combine the two summations. To do so, we will make their limits the same. We can modify the limits of the second summation from (r+1, x) to (1, x-r) as follows: $$ E[X]=(1-p) \sum_{i=1}^{r} p^{i-1}(i+E[X])+(1-p) \sum_{i=1}^{x-r} p^{i+r-1} r $$
03

3. Combine the two summations

Now that we have the same limits within the two summations, we can combine them into one summation: $$ E[X]=(1-p) \sum_{i=1}^{r} p^{i-1}(i+E[X]+rp^{r-1}) $$
04

4. Move the terms involving E[X] to the left

Now, in order to solve for E[X], we need to move all terms involving E[X] to the left of the equation. This results in: $$ E[X]\sum_{i=1}^{r}(1-p)p^{i-1}=(1-p) \sum_{i=1}^{r} p^{i-1}(i+rp^{r-1}) $$
05

5. Simplify the expected value expression

Now we can divide by the term \(\sum_{i=1}^{r}(1-p)p^{i-1}\) to obtain E[X] as follows: $$ E[X]=\frac{(1-p) \sum_{i=1}^{r} p^{i-1}(i+rp^{r-1})}{\sum_{i=1}^{r}(1-p)p^{i-1}} $$
06

6. Solve the summations to get the final expected value

The sum of a geometric series is given by the formula: $$ \sum_{n=0}^{N} ar^n = a\frac{1-r^{N+1}}{1-r} $$ Plugging in the values from our problem's summations and solving, we get: $$ E[X]=\frac{(1-p) \left( \frac{1-p^{r+1}}{1-p}\right) (1+rp^{r})}{\left( \frac{1-p^{r}}{1-p}\right)} = \frac{(1-p^{r})(1+rp^{r})}{1-p^{r}} = 1 + rp^{r} $$ So, the expected number of flips to obtain r heads in a row for a biased coin with probability p of heads is: $$ E[X]=1 + rp^{r} $$

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Key Concepts

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

Expected Value
The expected value is a fundamental concept in probability theory that represents the average outcome of a random event if it is repeated many times. It provides a measure of the center of a probability distribution. For a random variable, the expected value gives an idea of what one would expect the average result to be after many trials.

In the context of coin flips, expected value is used to determine how many times you might need to conduct an experiment—in this case, flipping the coin—before achieving a desired outcome, such as getting a certain number of heads in a row. Let's say that the coin lands on heads with probability \(p\). If you want to find out how many flips are needed on average to get \(r\) heads consecutively, you would calculate the expected value \(E[X]\) using specific probability formulas. In this exercise, the solution derived the expected value formula \(E[X]=1 + rp^{r}\), indicating that on average, 1 plus an additional term related to \(p\) and \(r\) flips are needed.
Geometric Series
A geometric series is a series with a constant ratio between successive terms. It's a critical tool in various mathematical calculations, including probability theory. The general form of a geometric series is \(a + ar + ar^2 + ar^3 + \ldots\), where \(a\) is the first term and \(r\) is the common ratio.

In probability exercises, geometric series help in calculating the total probability of multiple outcomes over repeated trials. In our exercise about coin flips, the geometric series formula \(\sum_{n=0}^{N} ar^n = a\frac{1-r^{N+1}}{1-r}\) was used to help solve the expected number of flips needed for obtaining heads \(r\) times consecutively. The simplification involves rearranging terms until they can be calculated using this compact form, which is crucial in getting manageable expressions for real-world problems.
Conditional Expectation
Conditional expectation is the expected value of a random variable given that another event has occurred. It refines the idea of expectation by narrowing it down to specific scenarios, adding another layer to probability calculation by considering additional information.

When calculating the expected number of flips in our coin example, conditional expectation played a vital role. By conditioning on when the first tails occurs, the problem is broken down into more manageable parts: assessing when tails will first appear within the sequence of flips. This involved understanding the probability under specific conditions, such as the number of flips until the first tails shows after \(r\) consecutive heads, and calculating the expected number for each case to reach the solution \(E[X]=1 + rp^{r}\). This method provides more detailed insight, allowing for complex probability scenarios to be evaluated systematically.
Biased Coin Flips
In probability theory, a biased coin is one that does not have an equal probability of landing on heads or tails; instead, it has a skewed outcome probability, represented as \(p\) for heads and \(1-p\) for tails.

Working with a biased coin requires adjusting your calculations to account for this unequal likelihood. When determining how many flips are expected to get \(r\) consecutive heads, one must consider this bias. Our exercise demonstrates how to handle this situation effectively using the derived expected value \(E[X]=1 + rp^{r}\), which incorporates the bias by including \(p\) directly in the formula. This calculation highlights the impact of bias in practical situations, showing how non-uniformity affects probabilistic predictions, and why accommodating such biases in computations is crucial to achieving accurate outcomes.

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