Probability Distribution Coin Toss

Introduction to probability distribution coin toss:

In probability, a probability distribution used to identify with in a particular interval falling value of a probability. The probability distribution describes the range of the possible value of a random variable. The probability distribution is entirely described by the cumulative distribution function. In this article we shall discuss probability distribution coin toss example problem.

Probability Distribution Coin Toss Example Problem

Example: A coin is tossed for 65times. Find the expectation value of tail with the help of using the binomial distribution.

Solution:

Consider 65 times the coin is tossed, so n = 65

When toss a coin, we get probability of tail is p = 1/2

Probability of getting tail is represent by q

q = 1 – p                        [p + q = 1]

q = 1 – 1/2

q = 1/2

Expectation value:

μ = E[x] = np

= 65(1/2 )

= (65/2)

μ = E[x] = 32.5

Answer:

Expectation value = μ = E[x] = 32.5

Example 2: An unbiased die is rolling for 45 times. By using the binomial distribution find the expectation value of getting four?

Solution:

Let p be probability of getting success

Let q = 1 – p probability of getting failure

Let n be number of times the dice is thrown

An unbiased dice is rolling for 45 times, the value of n = 43

When we throe a unbiased dice, probability of getting four is p = 1/6

Probability of not getting four is represent by q

q = 1 – p

= 1 – 1/6

= 6/6– 1/6

= 6-1/6

q = 5/6

Expectation value:

μ = E(x) = np

= 45 (1/6)

= 45/6

= 7.5

μ = E(x) = 7.5

Answer:

Expectation value = μ = E[x] = 7.5

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Probability Distribution Coin Toss Practice Problem

Problem:

A coin is tossed for 42 times. Find the expectation value of tail with the help of using the binomial distribution.

Answer:

Expectation value = 21

Problem:

An unbiased die is rolling for 35 times. By using the binomial distribution find the expectation value of getting four?

Answer:

Expectation value = 5.8

Mixed Probability Distribution

Introduction to mixed probability distribution:

In this article we shall discuss the mixed probability distributions. Probability plays the main part in the mathematics and in all new fields. Probability is used to discover number of incidence of an event out of total outcomes. The probability of certain two random variables X and Y define on the equivalent probability space, the mixed distribution for X and Y defines the probability of events distinct in expressions of together X and Y.

Basic for Mixed Probability Distribution:

We found with a random test among probability measure P on a primary sample space. Now, we will confer two "mixed" cases for the probability distribution of a random variable in this case where the distribution is partly discrete and partly continuous and the case where the variable has both discrete coordinates and continuous coordinates.

Mixed Probability Distributions Consist of Following Cases:

Cumulative distribution case:

The cumulative distribution function for a couple of random variables is distinct in terms of their mixed probability distribution,

`F(x,y)=P(Xlt=x,Ylt=y)` .

Discrete distribution case:

For discrete random variables, the mixed probability distribution is,

`P(X=x and Y=y)=P(Y=y|X=x).P(X=x)` .

`=P(X=x|Y=y).P(Y=y)` .

Since these are probabilities, we have

`sum_(x)` `sum_(y)`` P(X=x and Y=y)=1` .

Continuous distribution case:

Likewise for continuous random variables, the mixed probability distribution function can be denoted as fX,Y(x, y) and this is,

`f_(X.Y)(x,y)=f_(Y|X)(y|x)f_(X)(x)=f_(X|Y)(x|y)f_(Y)(y)` .

Anywhere fY|X(y|x) and fX|Y(x|y) provide the conditional distribution of Y known X = x and of X known Y = y in that order, and fX(x) and fY(y) give the secondary distribution for X and Y correspondingly. Is this topic math 4th grade word problems hard for you? Watch out for my coming posts.

Again, since these are probability distributions, one has

`int_(x)int_(y)f_(X,Y)(x,y)dydx=1` .

Mixed probability distribution case:

Most of time X is continuous but Y is discrete. Intended for case, in a logistic regression, one might wish to predict the probability of a binary outcome Y conditional on the value of a continuously-distributed X. On the other hand, a "mixed probability distribution" can be distinct in together of two ways.

`f_(X,Y)(x,y)=f_(X|Y)(x|y)P(Y=y)` .

`=P(Y=y|X=x)f_(X)(x)` .

Properly, fX,Y(x, y) is the probability mass function of (X, Y) with value to the product measure on the particular supports of X and Y. each of these two decompositions can then be used to pick up the varied cumulative distribution function.

`F_(X,Y)(x,y)=sum_(tlt=y)int_(s=-oo)^x f_(X,Y)(s,t)ds` .

Grouping of arbitrary statistics of discrete and continuous random variables.

Types of Data Distribution

Introduction to types of data distribution:
The data distribution is also said to be a theoretical distribution. There are three types of data distribution given below. Random variables are distributed regarding to probability law corresponding to probability distribution is called theoretical distribution. Here we are going to discuss about types of data distribution with suitable example problem.

Types of Data Distribution:

There are three types data distribution, such as

Binomial distribution
Normal distribution
Poisson distribution

Example Problems – Types of Data Distribution:

Example problem 1 – types of data distribution:

Obtain the given Poisson distribution where, e = 2.718, x = 10  and `lambda` = 7, x = 10

Solution:

Given: `lambda` = 7; x = 10

Formula to find Poisson distribution = `((e^(-lambda))(lambda^(x)))/(x!)`

First we find the value of `e^(-lambda)` .

e-3 = (2.718)-7

= 0.000911

Then you will find` lambda^(x)`

Given: `lambda` = 7; x = 10

Therefore, `lambda^(x) = ` `7^(10)`

= 282475249

Then you will substitute the values in given formula

`((e^(-lambda))(lambda^(x)))/(x!)` ` = (0.000911(282475249))/(10!)`

= `257334.951839/3628800`

= 0.07091

Answer: Poisson distribution = 0.07091

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Example problem 2 – Types of data distribution:

Obtain the value of p. In a Binomial distribution if n = 4 and P(r = 3) = 2P(r = 2).

Solution:

P(X = r) = nCr pr qn−r

P(X = 3) = 4C3 p3q1 and P(X = 2) = 4C2 p2q2

∴ 4C3 p3q1 = 2 (4C2 ) p2q2

4 p3q1 = 2 * 6 * p2q2

∴ 4 p = 12q

p = 3 q

p = 3 (1 − p) ⇒ 4p = 3; p =`3/4.`

Answer: The value of p = `3/4`

Example problem 3 – types of data distribution

Find the normal distribution of P( X `lt` 200) if the normal random variable has mean(m) = 150, standard deviation(`sigma` )  = 30.

Solutions:

Step 1: Given:

Mean (m) = 150

Standard deviation (`sigma` ) = 30

Step 2: Formula:

X `lt`  mean = 0.5 - Z

X `gt` mean = 0.5 + Z

X = mean = 0.5.

Z = `(X-m)/(sigma)`

Step 3: Solve:

Z = `(X-m)/(sigma)`

= `(200 - 150) / 30`

= `50/30`

= 1.67

Step 3: Find Z = 1.67, Refer Z table,

Z = 1.67 = 0.4525

Step 4: Solving:

X value greater than mean

P(X) = 0.5 + 0.4525

= 0.9525

Result: Normal Distribution P( X `lt` 200) = 0.9525

Solve by Substitution Solver

Introduction to solve by substitution solver:

Generally the solver uses either a elimination method, the matrix method, the substitution method, or the graphing method to solve the system of simultaneous equations. One of the easiest methods is solving simultaneous equations by substitution method. I like to share this How to do the Substitution Method with you all through my article.

Solving steps used by the substitution solver:

Step 1: Let us take any one of the equation. From this equation, write one variable in terms of the other variable.

Step 2: Next, substitute this in another one equation to get the single variable equation. Simplify and then find a one variable value.

Step 4: Substitute this variable value in any one of the equation, so we get the other another one variable value.

Example Problem to Solve by Substitution Solver:

Example problem 1:

Solve the following two simultaneous equations by substitution method:

x + y = 2---------Equation (1)

x – y = 4---------Equation (2)

Solution:

Step 1: Let us consider the equation (1)

x + y = 2

Subtract x on both sides of the equation

x + y - x = 2 – x

y=-x + 2------------------------Equation (3)

Step 2: Substitute this value of y in Equation (2). We get

x – y = 4

x – (-x + 2) = 4

x + x – 2 =4

2x – 2 = 4

Add 2 on both sides of the equation

2x – 2 + 2 = 4 + 2

2x = 6

Divide by 2 on both sides of the equation

2x/2 = 6/2

x = 3

Step 3: Plugging this value of x in Equation (3), we get

y=-x + 2

y=-3 + 2

y = -1

So, the solution of the simultaneous equations is (3, -1). Is this topic Derivatives Calculus hard for you? Watch out for my coming posts.

Additional Problem to Solve by Substitution Solver:

Solve the following two simultaneous equations by substitution method:

y = 2x ------------Equation (1)

x + 2y = 5------------Equation (2)

Solution:

Step 1: Let us consider the equation (1)

y = 2x

Step 2: Substitute the value of y in Equation (2). We get

x + 2y = 5

x + 2(2x) = 5

x + 4x = 5

5x = 5

Divide by 5 on both sides of the equation

5x/5 = 5/5

x = 1

Step 3: Plugging this value of x in Equation (1), we get

y= 2x

y = 2(1)

y = 2

So, the solution of the simultaneous equations is (1, 2).

Linear Equation in One Variable Definition

Introduction to linear equation in one variable definition:

The linear equations are the equations which contain one or more variables which contains the order 1. The linear equation is to find the roots for the given linear equations. The linear equation has only the simple terms for their equation. The linear equations of the two variables are calculated by the elimination method, substitution method and the graphing method. In this, first find any one of the values for the linear equations. Using that value find the value for other linear equation. Solve the linear equation is to find the values for the given linear equations. This article has the information about the linear equation in one variable definition.

Examples for Linear Equation in One Variable Definition:

Example 1 to linear equation in one variable definition:

Find the value for the linear equation 12x+ 14 = 566.

Solution:

The given linear equation is 12x+ 14 = 566.

Step 1: 12x+ 14 = 566

Step 2: 12x+14 - 14 = 566 - 14

Step 3: 12x = 552

Step 4: `(12x)/12` = `552/12`

Step 5: x = 46

The value for the linear equation 12x+ 14 = 566 is x = 46.

Example 2 to linear equation in one variable definition:

Find the value for the linear equation 10x+ 24 = 746.

Solution:

The given linear equation is 10x+ 24 = 746.

Step 1: 10x+ 24 = 746

Step 2: 10x+24 - 24 = 746 - 24

Step 3: 10x = 722

Step 4: `(10x)/10` = `722/10`

Step 5: x = 72.2

The value for the linear equation 10x+ 24 = 746 is x = 72.2.

Example 3 to linear equation in one variable definition:

Find the value for the linear equation 18x+ 17 = 852.

Solution:

The given linear equation is 18x+ 17 = 852.

Step 1: 18x+ 17 = 852

Step 2: 18x+17 - 17 = 852 - 17

Step 3: 18x = 835

Step 4: `(18x)/18` =` 835/18`

Step 5: x = 46.39

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Practice Problems for Linear Equation in One Variable Definition:

Find the value for the linear equation 2x+ 7 = 157.
Answer: x = 75

Find the value for the linear equation 5x+ 18 = 248.
Answer: x = 46

Operations in the Set Theory

One of the theories used mathematics is the set theory. There are many operations available in it. There can be intersection and union. These are just two of the operations but there are many more. Union just gives a set containing all the elements present in the components.  The intersection of set gives a new set which contains the common elements of the components. There can intersection of three sets as well. In this the common elements of all the three are included in the final answer. Once this is found out the task becomes simpler as all common elements are known.

A convex set can also be defined in this concept. This is very closely related to set theory. In fact it is a part of the set theory. It is better to understand this concept. The intersection operation can take place on these also. The intersection of complex sets denotes all the points which will be available on the line which will be joining two points in vector space. The intersection of sets definition math can be explained with examples. The examples can make the concepts clearer. The examples are always useful in explaining the concepts in mathematics. The intersection operation basically helps in finding the common elements. So, this operation can be very helpful in finding the common elements. The other operations can also be very helpful in the set theory. The set theory has got a concept called Venn diagrams. They can be very helpful and explanatory. They can be used to solve the problems very easily. They are diagrammatically representation of the operations in the set theory. It is always better to use the diagrams as they can make the concepts clear and easily understandable.Having problem with how to round decimals keep reading my upcoming posts, i will try to help you.

So, to solve the problems in the set theory it is better to use the Venn diagrams. Venn diagrams can make the complex problems in mathematics into simpler ones. Once the problems are made simple, they can be solved. There can be various intersections like arbitrary intersection and nullary intersection. Null set is a set which does not contain any element. Universal set is the set which is the master set. There is one more important concept in the set theory. This is the concept regarding the disjoint set. In intersection there are common elements. But in the case of disjoint set there are no common elements. So, the answer is a null set.

Primary Number Line

Introduction to primary number line:

The line is marked with the number is called primary number line. The number line has both positive and negative number. The center point of the primary number line is zero. The zero will not be neither positive nor negative. The positive may or may not have the positive sign on primary number line. But the negative numbers should have negative sign(“-“)

Example for Primary Number Line:

Example 1:

Point the expression x < 3 and x > -2 on the primary number line

Solution:

The expressions x < 3 and x > -2 can be written as -2
Example 2:

Point the expression n > -1 and n < 4on the primary number line

Solution:

The expressions n > -1 and n < 4on can be written as -1 < n < 4. This means, n is greater than -1 but less than 4. Please express your views of this topic second order differential equation by commenting on blog.

Practice Problems for Primary Number Line:

1. Point the expression x < 3 and x > -2 on the primary number line

2. Point the expression x < 5 and x > -5 on the primary number line

3. Point the expression x < 4 and x > -7 on the primary number line

5. Point the expression x < 2 and x > -8 on the primary number line

6. Point the expression x < 9 and x > -11 on the primary number line

7. Point the expression x < 12 and x > -1 on the primary number line

8. Point the expression x < 13 and x > -15 on the primary number line

9. Point the expression x < 4 and x > -8 on the primary number line

10. Point the expression x < 9 and x > -1 on the primary number line