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Chapter 3: Problem 24

A coin, having probability \(p\) of landing heads, is continually flipped until at least one head and one tail have been flipped. (a) Find the expected number of flips needed. (b) Find the expected number of flips that land on heads. (c) Find the expected number of flips that land on tails. (d) Repeat part (a) in the case where flipping is continued until a total of at least two heads and one tail have been flipped.

Short Answer

Expert verified
The short answers to each part of the question are: (a) The expected number of flips needed until at least one head and one tail have been flipped is \(E(N) = \frac{1}{2p(1-p)}\). (b) The expected number of flips that land heads is \(\frac{1}{2(1-p)}\). (c) The expected number of flips that land tails is \(\frac{1}{2p}\). (d) The expected number of flips needed until at least two heads and one tail have been flipped is \(E(N) + E(M) = \frac{1}{2p(1-p)} + \frac{1}{p}\).

Step by step solution

01

(a) Expected number of flips needed

We can denote the expected number of flips needed as E(N). We can also split this into two cases: the first flip is heads (with probability p) and the first flip is tails (with probability 1-p). After the first flip, we are essentially waiting for the other side to appear. So, we can write the expected value as: E(N) = p(E(N | First flip is heads)) + (1-p)(E(N | First flip is tails)) Now, let's consider the two conditional expected values: E(N | First flip is heads) = 1 + (1-p)E(N), since the first flip is heads and then we expect to need (1-p) more flips before getting a tail. E(N | First flip is tails) = 1 + pE(N), since the first flip is tails and then we expect to need p more flips before getting a head. Now we can plug these values back into the original equation: E(N) = p(1 + (1-p)E(N)) + (1-p)(1 + pE(N)) Now we can solve for E(N): E(N) = 1 + p(1-p)E(N)+ (1-p)(pE(N)) E(N)(1 - p(1-p) - (1-p)p) = 1 E(N) = \(\frac{1}{1 - p(1-p) - (1-p)p}\) E(N) = \(\frac{1}{2p(1-p)}\)
02

(b) Expected number of flips that land heads

Now, let's consider the ratio of heads to total flips: Ratio of heads = \(\frac{p \cdot E(N)}{E(N)}\) Expected number of flips that land heads = p * E(N) = \(\frac{p}{2p(1-p)}\) = \(\frac{1}{2(1-p)}\)
03

(c) Expected number of flips that land tails

Similarly, the expected number of flips that land tails can be calculated as: Expected number of flips that land tails = (1-p) * E(N) = \(\frac{1-p}{2p(1-p)}\) = \(\frac{1}{2p}\)
04

(d) Expected flips until two heads and one tail

Now, we need to find the expected number of flips needed to get at least two heads and one tail. Let E(M) be the expected number of flips needed to get one more head after getting one head and one tail. E(M) = p(1) + (1-p)(1 + E(M)) E(M) = \(\frac{1}{p}\) Now we can find the expected number of flips needed for two heads and one tail case: Expected number of flips = E(N) + E(M) = \(\frac{1}{2p(1-p)} + \frac{1}{p}\)

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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 models and statistics. It signifies the average outcome of an experiment or process that you would expect if you repeated it many times.
In the context of this exercise, you are asked to find the expected number of flips needed until certain conditions are met, like getting one head and one tail.
  • To calculate the expected number of flips needed, we consider each flip's probability and the possible outcomes.
  • The concept of expected value leverages conditional probabilities, which adjust expectations based on previous results.
This is mathematically represented using an equation that averages the possible outcomes weighted by their probabilities. Understanding expected value not only helps in homework solutions but also in real-world decision making, as it gives insight into what to typically expect in uncertain scenarios.
Coin Flipping
Coin flipping is a classic example used to illustrate probability models. It's simple, yet rich enough to explores complex probability topics.
Here, the coin may be biased, meaning it doesn't necessarily have a 50-50 chance for heads or tails:
  • Probability of landing heads is denoted by \( p \).
  • Probability of landing tails is \( 1-p \).
Coin flipping is ideal for demonstrating conditional probability because each flip's outcome may affect the expected results of subsequent flips. In our exercise, understanding the basic coin flipping mechanic allows us to assess further calculations on the expected flips needed, or on expected flips that result in heads or tails.
Conditional Probability
Conditional probability refers to the probability of an event occurring given that another event has already happened.
This is critical when considering the number of coin flips needed until certain outcomes:
  • We break down the expected scenarios by whether the first flip is heads or tails.
  • Once the first flip occurs, the expected number of additional flips changes based on the first result.
This exercise's solutions use conditional probability to refine calculations after every flip. This approach helps to systematically solve problems by narrowing down the potential outcomes based on already-known information.
Probability Distribution
Probability distribution gives a complete picture of all the possible outcomes of a random experiment and their associated probabilities. In the context of flipping a coin, it's a way to represent the entire range of outcomes.
  • Discrete probability distributions are appropriate because coin flips result in finite, countable outcomes (heads or tails).
  • Each flipping sequence ending in the desired conditions represents a different scenario in our distribution.
When solving for expected values using probability distributions, you effectively average across all outcomes, weighted by their probability. This distribution is fundamental in computing the expected values of the number of flips needed or the number of heads in those flips. Embracing how probability distributions work will strengthen your problem-solving mechanics in probability and statistics.

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Most popular questions from this chapter

There are three coins in a barrel. These coins, when flipped, will come up heads with respective probabilities \(0.3,0.5,0.7\). A coin is randomly selected from among these three and is then flipped ten times. Let \(N\) be the number of heads obtained on the ten flips. Find (a) \(P(N=0\\}\). (b) \(P\\{N=n\\}, n=0,1, \ldots, 10\) (c) Does \(N\) have a binomial distribution? (d) If you win \(\$ 1\) each time a head appears and you lose \(\$ 1\) each time a tail appears, is this a fair game? Explain.

Let \(X\) be exponential with mean \(1 / \lambda\); that is, $$f_{X}(x)=\lambda e^{-\lambda x}, \quad 01]\).

An urn contains \(n\) white and \(m\) black balls which are removed one at a time. If \(n>m\), show that the probability that there are always more white than black balls in the urn (until, of course, the urn is empty) equals \((n-m) /(n+m)\). Explain why this probability is equal to the probability that the set of withdrawn balls always contains more white than black balls. [This latter probability is \((n-m) /(n+m)\) by the ballot problem.]

A set of \(n\) dice is thrown. All those that land on six are put aside, and the others are again thrown. This is repeated until all the dice have landed on six. Let \(N\) denote the number of throws needed. (For instance, suppose that \(n=3\) and that on the initial throw exactly two of the dice land on six. Then the other die will be thrown, and if it lands on six, then \(N=2\).) Let \(m_{n}=E[N]\). (a) Derive a recursive formula for \(m_{n}\) and use it to calculate \(m_{i}, i=2,3,4\) and to show that \(m_{5} \approx 13.024\). (b) Let \(X_{i}\) denote the number of dice rolled on the \(i\) th throw. Find \(E\left[\sum_{i=1}^{N} X_{i}\right]\).

A and \(B\) roll a pair of dice in turn, with \(A\) rolling first. \(A\) 's objective is to obtain a sum of 6 , and \(B\) 's is to obtain a sum of 7 . The game ends when either player reaches his or her objective, and that player is declared the winner. (a) Find the probability that \(A\) is the winner. (b) Find the expected number of rolls of the dice. (c) Find the variance of the number of rolls of the dice.
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