Looking For The Best Statistics Homework Help?

The term statistics brings a lot of horrific memories to anyone who has gone through campus. A statistics class was not the best during the old college days. Today the trend is still on. Actually many students are not in a position to handle any homework in statistics. Many seek for statistics homework help from online platforms and sometimes friends who did the course. Such statistics homework help comes at a very high cost. Actually, there are professionals who have specifically specialized in statistics and offer statistics homework help to any student who finds it hard to grasp the head or tail of statistics. Such professionals make some good money out of such deals of helping students. There is also the other category of individuals who are specialist of a given statistics topic. A good example is those who offer calculus homework help. For ages up to date, statistics have remained a course for the elite few. It is a course set aside for bookworms who often spend most of their time in the library going through a lot of statistics books.

Statistics have different topics. The main topics are calculus, probability, hypothesis testing, time series analysis just to mention but a few. Many students will seek for statistics homework help in many of these topics. Probability often tops the most sought for homework help. It is followed closely by those who seek calculus homework help. Actually many are times students seeking help solving statistics problems will be in the same numbers both for calculus and probability problems. The rest of the topics are also on high demand among those seeking homework help but not as much as the main two. A large number of students seeking help solving statistics problems has attracted so many people who want to offer homework help. Some of the ones who purport to offer homework help services are conmen. There is a number of students who have found themselves victims of such con men. Many students have suffered a double loss when they have been conned since they do not make it submit the assignments. It can all be explained by the fact that many students who seek for statistics homework help do not have any idea on the homework and therefore they give up when conned. As a result, they do not submit any task to the tutor. Many will end up having missing marks and they have to redo the unit before they graduate.

Solving statistics problems has also gone online. There are many statistics online tutors lined up in different online platforms to offer math homework help. Interestingly calculus homework solutions online are sought more in comparison to other statistics topics. The main reason for this is the high flexibility and accessibility of getting online homework help. Besides online homework tutors tend to offer pocket-friendly services compared to their friends in cybers and other areas that offer statistics homework solutions.

Statistics homework help that we provide includes probability and calculus which are ones that are enormously sought for as far as seeking homework help is concerned. Calculus is such a dynamic topic. It involves mainly differential and integrals. Calculus grows algebra whereas algebra grows arithmetics. That might sound like rocket science. However, for now, let's look at probability in our discussion and probably have some basics that can kick start any student who seeks for statistics homework help in the field of probability.

We can not consider probability without defining some terms. The terms are an event, sample space, and experiment. The experiment is the activity done to result in an event. Sample space is all the possible outcomes. The event is a collection of possible outcomes.

Example

Let's consider rolling a dice.

The act of tossing the dice is the experiment

The sample space (1,2,3,4,5,6)

The event is either Odd (1,3,5) or Even (2,4,6)

Example 2

If you toss a fair coin three times. What is the possibility of getting a Head at least ones?

Possible outcomes

( HHH, THH, HTH, HHT, HTT, THT, TTH, TTT )

Let's have some definition

The probability of having no H is Event X

The probability of having one H is Event Y

The probability of having at least one H is Event Z

The probability of having no T is Event A

The probability of having exactly one T is Event B

The probability of having more than one T is Event C

Kindly note that each statement is about the probable outcomes that each event will produce.

Event X = no H = TTT

Event Y = getting exactly one H =

HTT, THT, TTH,

Event Z = getting at least one H =

HTT, THT, TTH, THH, HTH, HHT, HH

A = no T = HHH

Event B = outcome being exactly one T =

THH, HTH, HHT

Event C =

Outcome being at least one T

= THH, HTH, HHT, HTT, THT, TTH, TTT

From the above, you can now comprehend more about probability. Let's move on with our discussion.

Other definitions

Mutually exclusive events.

The events cannot occur at the same time. A choice is made between one of them.

For example.

When choosing which of the two shirts you will wear. Certainly, you can not wear both of them at ago, you will have to wear one of them.

Another example if you are buying a car you can not buy a car with both golden seats and silver seats. You will have to choose one of them.

Besides, you cannot be walking at the same time you are sitting.

The events are then termed as mutually exclusive.

Mutually inclusive events.

They can happen at the same time. You can do both of them at ago. Mostly consist of events that occur hand in hand. The events cannot happen independently of each other.

For example.

When you toss a dice. Certainly, if 6 is at the top 1 will be in the bottom.

You can walk and talk at the same.

When you drive a car you must have a driving license.

If you are using electricity then there will be an electric bill to clear at the end of the day.

Addition rule

The rule is used to compute the probability of either two events happening

It is used to calculate the probability of both mutually inclusive events as well as that of mutually exclusive ones.

Formula.

P ( A or B)

= P ( A) + P ( B) - P ( A & B)

Example.

If you have a normal number of cards. Find the chances of either picking a queen or a heart.

It is possible to get a queen and a heart.

P ( getting a queen or a heart) = P (heart) + P ( queen)- P (queen & heart)

= 4/52 + 13/52 - 1/52

= 16/52

Rule of subtraction

Two clear points that you need to note.

- The probability of an event ranges from 0 to 1.

-The sum of all possible events equals 1.

The rule of subtraction follows the above points.

The probability of event A occurring equals one minus the probability of event A not occurring.

P ( A )= 1-P ('A)

Example.

If the probability of John graduating from high school is 0.60. What is the probability of him not graduating?

P ( not graduating )

= 1- P ( graduating )

= 1- 0.60

= 0.40

Example 2

The probability of a given candidate passing an election is 30%. What is the probability that the candidate will not pass the election?

P ( not pass election )

= 1 - P ( pass election )

= 1- 0.30

= 0.70

Multiplication rule

It is used when you want to find out the chances of both events occurring.

The probability that both A and B occurs equals the probability of event A occurring multiple by the probability of event B occurring given that A has occurred.

P ( A & B )

= P ( A ) * P ( B | A)

Example 1

A dish contains 6 red balls and 4 black balls. Two balls are drawn from the dish without replacement.

What are the chances that both balls are black?

Let A be the first ball is black

Let B be second marble is black

At first, the probability that first balls are black P ( A ) = 4/10 since there are 4 black balls

When you draw the first ball there are 3 black balls remaining of the total 9

P ( B ) = 3/9

The probability that both balls are black

P ( A & B )

= P ( A) * P ( B | A)

= 4/10 * 3/9

= 12/90

= 2/15

Example 2

A bag has 20 marbles. 5 of them are white, 8 are green and 7 yellow. Three balls are drawn without replacement.

What are the chances that all the balls drawn are white?

Let X be the first white ball

Let Y be the second white ball

Let Z be the third white ball

The probability that the first ball is white

P ( X ) = 5/20

The probability that the second ball is white. There are 4 white balls and 19 balls in total.

P ( Y ) = 4/19

The probability that the third ball is white. There are 3 white balls and 18 balls in total.

P ( Z ) = 3/18

Probability that all balls are white

P ( X Y Z )

= P ( X ) * P ( Y ) * P ( Z )

= 5/20 * 4/19 * 3/18

= 60 / 6840

= 0.00877

Conclusion.

Probability is one of the many topics in statistics. If you are looking for help solving statistical problems in the area of probability then you do not need to go far. It is all laid out in the discussion of the article. Consider the information outlined here before you look for those statistics homework help. You can do it.

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