/*! 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 16 Let \(Q_{n}\) denote the probabi... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(Q_{n}\) denote the probability that in \(n\) tosses of a fair coin no run of 3 consecutive heads appears. Show that $$ \begin{aligned} &Q_{n}=\frac{1}{2} Q_{n-1}+\frac{1}{4} Q_{n-2}+\frac{1}{8} Q_{n-3} \\ &Q_{0}=Q_{1}=Q_{2}=1 \end{aligned} $$ Find \(Q_{8}\). HINT: Condition on the first tail.

Short Answer

Expert verified
The probability of no run of 3 consecutive heads in 8 tosses of a fair coin, denoted by \(Q_8\), is \(\frac{405}{512}\).

Step by step solution

01

Verify the given expression

At every stage, there are 2 possibilities: heads (H) or tails (T). Let's analyze the \(n\)-th toss and the possibilities that can prevent a run of 3 consecutive heads from appearing. 1. The \(n\)-th toss is tails: \(T\) We don't need to worry about the run of 3 consecutive heads in this case, and we just focus on the previous \((n-1)\) tosses. So the probability for this situation is \(Q_{n-1}\) multiplied by the probability of getting tails: \[\frac{1}{2} Q_{n-1}\] 2. The \(n\)-th toss is heads: \(H\) If we get heads in the \(n\)-th toss, there are 2 possibilities in the \((n-1)\)-th toss to avoid 3 consecutive heads: a. The \((n-1)\)-th toss is tails: \[\frac{1}{2} Q_{n-2}\] b. Both \((n-1)\) and \((n-2)\)-th tosses are heads, and the third last toss is tails: \[\frac{1}{8} Q_{n-3}\] Now, we sum the probabilities for all possible scenarios: \[Q_n = \frac{1}{2} Q_{n-1} + \frac{1}{4} Q_{n-2} + \frac{1}{8} Q_{n-3}\]
02

Calculate the base probabilities

Given base values: \[Q_0 = Q_1 = Q_2 =1\] We can use these to calculate the next probability value: \[Q_3 = \frac{1}{2} Q_{2} + \frac{1}{4} Q_{1} + \frac{1}{8} Q_{0} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8}\]
03

Calculate Q_8 recursively

Using the same approach, we can calculate the probability up to \(Q_8\): \[Q_4 =\frac{1}{2} Q_{3} + \frac{1}{4} Q_{2} + \frac{1}{8} Q_{1} = \frac{1}{2}\cdot\frac{7}{8} + \frac{1}{4}\cdot1 + \frac{1}{8}\cdot1 = \frac{29}{32}\] \[Q_5 =\frac{1}{2} Q_{4} + \frac{1}{4} Q_{3} + \frac{1}{8} Q_{2} = \frac{1}{2}\cdot\frac{29}{32} + \frac{1}{4}\cdot\frac{7}{8} + \frac{1}{8}\cdot1 = \frac{57}{64}\] \[Q_6 =\frac{1}{2} Q_{5} + \frac{1}{4} Q_{4} + \frac{1}{8} Q_{3} = \frac{1}{2}\cdot\frac{57}{64} + \frac{1}{4}\cdot\frac{29}{32} + \frac{1}{8}\cdot\frac{7}{8} = \frac{111}{128}\] \[Q_7 =\frac{1}{2} Q_{6} + \frac{1}{4} Q_{5} + \frac{1}{8} Q_{4} = \frac{1}{2}\cdot\frac{111}{128} + \frac{1}{4}\cdot\frac{57}{64} + \frac{1}{8}\cdot\frac{29}{32} = \frac{213}{256}\] \[Q_8 = \frac{1}{2} Q_{7} + \frac{1}{4} Q_{6} + \frac{1}{8} Q_{5} = \frac{1}{2}\cdot\frac{213}{256} + \frac{1}{4}\cdot\frac{111}{128} + \frac{1}{8}\cdot\frac{57}{64} = \frac{405}{512}\] Therefore, \(Q_8 = \frac{405}{512}\).

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Key Concepts

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

Recurrence Relations
Recurrence relations are equations that express each term of a sequence as a function of preceding terms. They serve a crucial role in solving problems involving sequences, particularly in probability theory. In our exercise, we have a recurrence relation for determining the probability of not having a run of three consecutive heads in a series of coin tosses. The general form given is:
  • \[ Q_n = \frac{1}{2}Q_{n-1} + \frac{1}{4}Q_{n-2} + \frac{1}{8}Q_{n-3} \]
This equation helps us calculate the probability for each toss based on the results of the previous three tosses. It reflects the different scenarios a coin toss could face: either a tail or a different combination of heads for the last few tosses. Specifically, it relies on prior probabilities, which are multiplied by the respective probabilities of each scenario.
This step-by-step dependency of each new probability value on preceding terms is a signature characteristic of recurrence relations in probability context. Such a framework simplifies complex probability calculations and enables us to solve for specific terms, like finding \( Q_8 \) in this case.
Combinatorial Probability
Combinatorial probability deals with finding the likelihood of an event contained within a defined set of possibilities, utilizing various counting techniques. In the context of our exercise, we apply these principles to examine the sequences of coin tosses to count valid configurations that avoid three consecutive heads. The formula used computes probabilities considering multiple outcomes:
  • The probability of having a tail on the \( n \)-th toss.
  • The probability of having a head on the \( n \)-th toss, given the prior two tosses avoid a consecutive run of heads.
Combinatorial analysis underlies the multiple scenarios contributing to \( Q_n \), requiring us to calculate separate outcomes and sum them. Understanding this approach helps in grasping how multiple outcomes are orchestrated in probability theory to build the complete picture of potential scenarios in various experiments, such as the given coin toss exercise.
Conditional Probability
Conditional probability refers to the probability of an event given that another event has occurred. In our exercise, it's used to break down complex events (like coin toss sequences without three consecutive heads) into manageable parts. By conditioning on the first toss, we isolate possible sequences based on its outcome, simplifying the calculations. For instance, if the first toss is a tail (\( T \)), we immediately set \( Q_{n} = \frac{1}{2}Q_{n-1} \). This simplifies because the risk of achieving three consecutive heads right from the start is nil. If the first toss is a head (\( H \)), we must further investigate conditions on the second and third tosses to ensure the absence of consecutive heads. By examining each condition separately:
  • The role of the second and third tosses impacts the resulting probability, as they parameterize further permissible formations.
  • Each step integrates the probability of no consecutive heads from prior tosses' configurations.
This systematic breakdown based on conditions enables us to construct precise probability calculations and reinforce the concept of dependent events in probability theory.
Coin Toss Experiment
The classic coin toss experiment is a fundamental illustration of probability theory. In our exercise, each coin toss acts as an independent trial, which collectively form a sequence. The simplicity of each toss offering a 50/50 chance (either heads \( H \) or tails \( T \)) becomes complex when conditions are added, such as avoiding consecutive heads. The task here is to evaluate the probability distribution over \( n \) coin tosses with an additional rule. The recurrence relation showcases how outcomes interlock over trials, calculating probabilities based on the occurrence or absence of heads in combinations:
  • The independence of each toss enriches the probability space, making predictions reliant on the past few outcomes.
  • Every new toss introduces a shared probability, reflecting its immediate predecessor tosses' outcomes as part of a rolling pattern.
Experimenting with these coin toss scenarios sharpens understanding of basic probability concepts, particularly the implications of increased trial numbers on probability distribution and outcome variability. This bedrock of probability theory is built on straightforward experiments like coin tosses, escalating as we analyze larger sequences and intricate conditions, as seen in the exercise's goal to find \( Q_8 \).

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