# 数学代写|概率论代写Probability theory代考|BERNOULLI TRIALS

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## 数学代写|概率论代写Probability theory代考|BERNOULLI TRIALS

Repeated independent trials are called Bernoulli trials if there are only two possible outcomes for each trial and their probabilities remain the same throughout the trials. It is usual to denote the two probabilities by $p$ and $q$, and to refer to the outcome with probability $p$ as “success,” $S$, and to the other as “failure,” $F$. Clearly, $p$ and $q$ must be non-negative, and
$$p+q=1 .$$
The sample space of each individual trial is formed by the two points $S$ and $F$. The sample space of $n$ Bernoulli trials contains $2^n$ points or successions of $n$ symbols $S$ and $F$, each point representing one possible outcome of the compound experiment. Since the trials are independent, the probabilities multiply. In other words, the probability of any specified sequence is the product obtained on replacing the symbols $S$ and $F$ by $p$ and $q$, respectively. Thus $\mathbf{P}{(S S F S F \cdots F F S)}=p p q p q \cdots q q p$.

Examples. The most familiar example of Bernoulli trials is provided by successive tosses of a true or symmetric coin; here $p=q=\frac{1}{2}$. If the coin is unbalanced, we still assume that the successive tosses are independent so that we have a model of Bernoulli trials in which the probability $p$ for success can have an arbitrary value. Repeated random drawings from an urn of constant composition represent Bernoulli trials. Such trials arise also from more complicated experiments if we decide not to distinguish among several outcomes and describe any result simply as $A$ or non- $A$. Thus with good dice the distinction between ace $(S)$ and non-ace $(F)$ leads to Bernoulli trials with $p=\frac{1}{6}$, whereas distinguishing between even or odd leads to Bernoulli trials with $p=\frac{1}{2}$. If the die is unbalanced, the successive throws still form Bernoulli trials, but the corresponding probabilities $p$ are different. Royal flush in poker or double ace in rolling dice may represent success; calling all other outcomes failure, we have Bernoulli trials with $p=\frac{1}{649,740}$ and $p=\frac{1}{36}$, respectively. Reductions of this type are usual in statistical applications. For example, washers produced in mass production may vary in thickness, but, on inspection, they are classified as conforming $(S)$ or defective $(F)$ according as their thickness is, or is not, within prescribed limits.

The Bernoulli scheme of trials is a theoretical model, and only experience can show whether it is suitable for the description of specified observations. Our knowledge that successive tossings of physical coins conform to the Bernoulli scheme is derived from experimental evidence. The man in the street, and also the philosopher K. Marbe, ${ }^2$ believe that after a run of scventecn heads tail becomes more probable. This argument has nothing to do with imperfections of physical coins; it endows nature with memory, or, in our terminology, it denies the stochastic independence of successive trials. Marbe’s theory cannot be refuted by logic but is rejected because of lack of empirical support.

## 数学代写|概率论代写Probability theory代考|THE BINOMIAL DISTRIBUTION

Frequently we are interested only in the total number of successcs produced in a succession of $n$ Bernoulli trials but not in their order.

The number of successes can be $0,1, \ldots, n$, and our first problem is to determine the corresponding probabilities. Now the event ” $n$ trials result in $k$ successes and $n-k$ failures” can happen in as many ways as $k$ lettcrs $S$ can be distributed among $n$ places. In other words, our event contains $\left(\begin{array}{l}n \ k\end{array}\right)$ points, and, by definition, each point has the probability $p^k q^{n-k}$. This proves the

Theorem. Let $b(k ; n, p)$ be the probability that $n$ Bernoulli trials with probabilities $p$ for success and $q=1-p$ for failure result in $k$ successes and $n-k$ failures. Then
$$b(k ; n, p)=\left(\begin{array}{l} n \ k \end{array}\right) p^k q^{n-k} .$$
In particular, the probability of no success is $q^n$, and the probability of at least one success is $1-q^n$.

We shall treat $p$ as a constant and denote the number of successes in $n$ trials by $\mathbf{S}_n$; then $b(k ; n, p)=\mathbf{P}\left{\mathbf{S}_n=k\right}$. In the general terminology $\mathbf{S}_n$ is a random variable, and the function (2.1) is the “distribution” of this random variable; we shall refer to it as the binomial distribution. The attribute “binomial” refers to the fact that (2.1) represents the $k$ th term of the binomial expansion of $(q+p)^n$. This remark shows also that
$$b(0 ; n, p)+b(1 ; n, p)+\cdots+b(n ; n, p)=(q+p)^n=1,$$
as is required by the notion of probability. The binomial distribution has been tabulated. ${ }^3$

# 概率论代考

## 数学代写|概率论代写Probability theory代考|BERNOULLI TRIALS

$$p+q=1 .$$

## 数学代写|概率论代写Probability theory代考|THE BINOMIAL DISTRIBUTION

$$b(k ; n, p)=\left(\begin{array}{l} n \ k \end{array}\right) p^k q^{n-k} .$$

$$b(0 ; n, p)+b(1 ; n, p)+\cdots+b(n ; n, p)=(q+p)^n=1,$$

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