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A coin is tossed repeatedly and, on each occasion, the probability of obtaining a head is \(p\) and the probability of obtaining a tail is \(1-p(0

Short Answer

Expert verified
(a) The probability is \(p^n\). (b) \(p_n = p^{n-1}(1-p)\). (c) The expected number is \(\frac{1}{1-p}\).

Step by step solution

01

Understanding the Problem

We have a coin with probabilities of landing heads (\(p\)) and tails (\(1-p\)) on each toss. We need to determine three things: the probability of never getting a tail in the first \(n\) tosses, the probability of getting the first tail on the nth toss, and the expected number of tosses to get the first tail.
02

Probability of No Tail in First n Tosses

If we want no tails in the first \(n\) tosses, then each toss must be a head. The probability of getting a head is \(p\). Therefore, the probability of getting all heads in \(n\) tosses is \(p^n\).
03

Probability of First Tail on nth Toss

To get the first tail on the \(n\)th toss, the first \(n-1\) tosses must be heads, and the \(n\)th toss must be a tail. The probability of the first \(n-1\) tosses being heads is \(p^{n-1}\), and the probability of the \(n\)th toss being a tail is \(1-p\). Therefore, the probability \(p_n\) is \(p^{n-1}(1-p)\).
04

Expected Number of Tosses for First Tail

The expected number of tosses to get the first tail follows a geometric distribution with success probability \(1-p\). The expected value for a geometric distribution is given by \(\frac{1}{q}\), where \(q\) is the probability of success in one trial. Here, the success is obtaining a tail, so \(q = 1-p\). Thus, the expected number of tosses is \(\frac{1}{1-p}\).

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Key Concepts

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

Geometric Distribution
Imagine you have a coin that you keep flipping until you get a tail. This problem is a classic example of a geometric distribution. In probability theory, a geometric distribution tells us the probability of the first occurrence of success. Here, the 'success' is getting a tail.

The geometric distribution has some key characteristics:
  • It deals with trials that are repeated until a success (tail) occurs.
  • Each trial is an independent event, meaning the result of one flip doesn't affect the next one.
  • The probability of success (a tail) remains constant in each trial, which is \(1-p\) in this scenario.
These characteristics help us understand the chance of first seeing a tail after a series of head tosses. Findings reference the geometric sequence here, illustrating the probability decreases exponentially with more head tosses initially, up until we get a tail.
Expectation in Probability
In probability theory, the expectation or expected value is a fundamental concept that predicts the average outcome of a random event over many trials. For geometric distributions, we use it to figure out how many attempts it might take to achieve our first success.

Here, we aim to find out the expected number of coin tosses before landing a tail. This involves understanding that, on average, with each trial, the probability of eventually hitting our first success can be calculated using the formula: \[\frac{1}{1-p}\]
This formula is derived from the nature of geometric distributions and provides a simple yet powerful insight. It boils down to saying: if the probability of success on a single trial is \(1-p\), how soon should you expect to see that first success? It shows us that as the likelihood of getting a tail (\(1-p\)) becomes smaller, the number of attempts needed will grow. The expected value is a vital tool for understanding how long we have to wait, on average, which is key in many real-life applications of probability.
Tossing Coins
Tossing coins is a classic example in probability theory due to its simplicity and clear outcomes: heads or tails.

When we toss a coin multiple times, each event represents an independent trial with two possible outcomes:
  • A head with probability \(p\).
  • A tail with probability \(1-p\).
Every toss is separate from the others, which means the outcome of one toss does not change the probabilities of the next.
Moreover, probability in tossing coins can be used to understand more complex ideas like the binomial and geometric distributions, as we see in exercises like these. The beauty of coin tossing problems lies in their straightforward setup, enabling us to leverage these ideas to tackle more complex and realistic situations, preparing us for the intricate world of probability and statistics. This fundamental grasp on simple experiments like coin tosses forms the foundation for deeper understanding and analysis in the field.

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Most popular questions from this chapter

Let \(X\) have distribution function \(F\). Find the distribution of \(Y=a X+b\) and of \(Z=|X|\).

An ambidextrous student has a left and a right pocket, each initially containing \(n\) humbugs. Each time she feels hungry she puts a hand into one of her pockets and if it is not empty, takes a humbug from it and eats it. On each occasion, she is equally likely to choose either the left or the right pocket. When she first puts her hand into an empty pocket, the other pocket contains \(H\) humbugs. Show that if \(p_{h}\) is the probability that \(H=h\), then $$ p_{h}=\left(\begin{array}{c} 2 n-h \\ n \end{array}\right) \frac{1}{2^{2 n-h}}, $$ and find the expected value of \(H\), by considering \(\sum_{h=0}^{n}(n-h) p_{h}\), or otherwise.

It is assumed that the number \(X\) of individuals in a population, whose fingerprints are of a given type, has a Poisson distribution with some parameter \(\lambda\). (a) Explain when and why this is a plausible assumption. (b) Show that \(\mathbf{P}(X=1 \mid X \geq 1)=\lambda\left(e^{\lambda}-1\right)^{-1}\). (c) A careless miscreant leaves a clear fingerprint of type \(t\). It is known that the probability that any randomly selected person has this type of fingerprint is \(10^{-6}\). The city has \(10^{7}\) inhabitants and a citizen is produced who has fingerprints of type \(t\). Do you believe him to be the miscreant on this evidence alone? In what size of city would you be convinced?

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An archer shoots arrows at a circular target of radius 1 where the central portion of the target inside radius \(\frac{1}{4}\) is called the bull. The archer is as likely to miss the target as she is to hit it. When the archer does hit the target, she is as likely to hit any one point on the target as any other. What is the probability that the archer will hit the bull? What is the probability that the archer will hit \(k\) bulls in \(n\) attempts? Prove that the mean number of bulls that the archer hits in \(n\) attempts is \(n / 32\). Show that if the archer shoots 96 arrows in a day, the probability of her hitting no more than one bull is approximately \(4 e^{-3}\). Show that the average number of bulls the archer hits in a day is 3 , and that the variance is approximately \((63 \sqrt{3} / 64)^{2}\).

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