Probability, binomial law: senior high school math course in PDF

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1 I. Proof, binomial distribution and Bernoulli’s scheme.
2 II. The binomial law.
3 III. Mind map.

Probabilities are applicable in everyday life. This chapter helps you to progress and solve problems with a new method. Probabilities are very interesting and you will encounter them everywhere. Follow the course on probability and the binomial distribution to fully understand this chapter.

I. Proof, binomial distribution and Bernoulli’s scheme.

1. test of Bernoulli.

Definition:

Let p be a real number belonging to [O ; 1].
A Bernoulli trial is any random experiment that has only two outcomes, called
generally success S and failure $\overline{S}$ and of respective probabilities and $q\,=\,1\,-\,p$ .

Bernoulli test

Examples:
– Tossing a balanced coin and finding out if tails is obtained is a Bernoulli trial of success S “Tails was obtained” whose probability is p = 0.5.
The failure S is” Face was obtained.
– Interviewing a person in the street in France and asking if they are drunk
is a Bernoulli test of success S “The person is left-handed.
probability is $p\,\simeq\,0,13$ .

2.Application and method on probabilities.

Example:

One card is drawn from a deck of 52 cards. For each of the following trials, indicate whether it is a Bernoulli trial and specify the success and its probability if applicable.
1. We look at the color of the card (spades, hearts, diamonds or clubs).
2. We check that the card is a spade.
3. We see if the card is not a spade.
4. We check that the card is an ace.
5. We look at the value of the card (ace, 2, 3, etc.).
6. We check that the card is a figure (king, queen or jack).

Solution:

1. Four outcomes are possible. This is not a Bernoulli test.
2. It is a Bernoulli test of success “The card is a spade” whose probability is $p\,=\,\frac{13}{52}=0,25$ .
3. It is a Bernoulli test of success “The card is not a spade” whose probability is $p\,=\,\frac{39}{52}=\,0,75$ .
4. It is a Bernoulli test of success ” The card is an ace ” whose probability is $p\,=\frac{4}{52}=\frac{1}{13}$ .
5. Thirteen outcomes are possible. This is not a Bernoulli test.
6. It is a Bernoulli test of success “The card is a figure” whose probability is $p\,=\frac{12}{52}=\frac{3}{13}$ .

3. the law of Bernoulli.

Definition:

One carries out a test of Bernoulli whose success S has probability p.
A random variable X is a Bernoulli random variable when it is at
values in {0; 1} where the value 1 is assigned to success.
We say then that X follows the Bernoulli distribution of parameter p.
In other words, we and .
We can summarize Bernoulli’s law by the following table.

Ownership:

Let X be a random variable that follows a Bernoulli distribution of parameter p.
The mathematical expectation of X is E(X) = p.

The variance of X is V(X)=p(1 -p).

Proof:
The expectation E(X) of X is :

$E(X)=P(X=1)\times \,1+P(X=0)\times \,0\,\\E(X)=p\times \,1+(1-p)\times \,0\,\\E(X)=p$

The variance V(X) of X is :

$V(x)=P(X=1)\,\times \,(1-E(X))^2+P(X=0)\times \,(0-E(X))^2$

$V(X)=p\times \,(1-p)^2+(1-p)\times \,(0-p)^2\,\\V(X)=p\,(1-p)^2+p^2\times \,(1-p)\,\\V(X)=p\,(1-p)(1-p+p)\,\\V(X)=p\,(1-p)$

4.4 Bernoulli diagram.

Definition:

Let n be a non-zero natural number.

A Bernoulli scheme is the repetition of n identical and independent Bernoulli trials.

Example:
Consider an opaque urn in which have been placed one green ball and two blue balls, all indistinguishable to the touch. And, we take a ball from this urn, we note its color, then we put the ball back in the urn. The experiment is repeated ten times and we are interested in the blue balls obtained. Each draw is a Bernoulli test of success S “The ball is blue” whose probability is $\frac{2}{3}$ .
As the ten draws are done with discount, the draws are identical and independent: we have a Bernoulli scheme.

II. The binomial law.

1. definition of the binomial distribution.

Definition:

Let n be a non-zero natural number and p a real number in the interval [O; 1].
We note X the random variable counting the number of successes obtained at the time of n identical and independent repetitions of a scheme of Bernoulli whose p is the probability of success.
We say then that X follows the binomial distribution of parameters n and p.

Ownership:

Let k be a natural number less than or equal to n and X a random variable that follows the
binomial law of parameters n and p.

Proof:

In a Bernoulli scheme, each path to k successes also allows
to obtain n- k failures. Each of these paths has the probability $p^k(1-p)^{n-k}$ .
Each path is determined by the data of its k successes: the number of paths
leading to k successes is equal to the number of combinations of k among n.
We deduce that :

Illustration with n = 3 :
In red the paths leading to k = 2 successes.

There are 3 ways to obtain two successes.

Each of them corresponds to a probability equal to .

2. Graphical representation of the binomial distribution.

The representation of the distribution corresponding to a binomial distribution depends on the parameter p: the closer p is to 0, the lower the probability of success.

If p becomes close to 1 then the probability of obtaining a large number of successes will be high.

Below, we see what happens with n = 8 and different values of p.
The height of each stick at x-axis k corresponds to P(X= k).

3. Expectation, variance and standard deviation.

Properties:

Let n be a strictly positive natural number and p a real number in the interval [0; 1].
We note X a random variable which follows the binomial distribution of parameters n and p.
1. The expectation of X is E(X)=np .
2. The variance of X is $V(X)\,=\,np(n-p)$ .
3. The standard deviation of X is $\sigma\,(X)=\sqrt{np(1-p)}$ .