In this tutorial, I am going to show you how to create a CSS animation in which the text “Neon Love” will appear on your webpage. Next, open up a new document in your text editor and type “Neon Love” in it. Now select all of the text with your mouse or keyboard and copy it. Paste the text into your code. *How to Make an Animation CSS Neon Love – CSS JavaScript Tutorial*

The next thing that you need to do is go to your html file and paste the code into it. Now go to your style sheet and add a class of “neon-love” to the text. Then set the text color to green and make sure that the font size is 12px. To make the text appear on your webpage, you will need to add this code:

Learn how to make a Java Script tutorial using neon colors. Find out how to make an animation of CSS Neon Love. How to make an animation neon light love in javascript. This tutorial will teach you how to make a neon light effect on a website using CSS.

*How to Make an Animation CSS Neon Love – CSS JavaScript Tutorial*

*How to Make an Animation CSS Neon Love – CSS JavaScript Tutorial*

And the probability of getting any number is 16. To find out the expected value, instead of adding all the numbers directly, multiply each number by the probability of getting that number. In this case the expected value will be 1⋅16+2⋅16+3⋅16+4⋅16+5⋅16+6⋅16=16(1+2+3+4+5+6)=3.5. That means if you play an infinite number of times you will get 3.5 on average each time. The thing to note here is that we have 3.5 in sixes

Where not written, it is the average value obtained by experimenting an infinite number of times.

Even if you think for yourself without the math, you can easily understand the matter, 3.5

are 1 to 6

The exact middle number, when the probability of getting any number is equal, it is normal to get the middle number if played infinitely.

expected

Now suppose that for some reason some sides of the six are heavy and some sides are light, so the probability of getting each number is not the same. The probability of getting x each time it is thrown is p(x) and of course p(1)+p(2)+…+p(6)=1. So what will be the expected value? Very simple, last time you multiplied 16 with each number, now multiply p(x) with x. Expected value is then E=p(1)∗1+p(2)∗2+…+p(6)∗6

.

We are trying to find a simple formula that will always work. If the six does not have 6 sides then n

What will happen if the side? Then the word six can not be used, say dice! But there will not be much change in the expected value formula, now we will add n terms. E=p(1)∗1+p(2)∗2+…+p(n)∗n=∑ni=1p(i)⋅i

.

We can make further generalizations, so long as we hold n

Each number from 1 to n appears once on the ta side. Now let us assume that the number written on the i-th side of the dice is x(i) and the probability of getting the i-th side after throwing the dice is p(i). Now the formula will be E=p(1)∗x(1)+p(2)∗x(2)+…+p(n)∗x(1)=∑ni=1p(i)⋅x(i). Simply put, for every i we call the number written on the i-th side as i

Multiplying by the probability of getting the th side and adding all the products.

When you read any academic book on mathematics, you will see that the above formula is given in the definition of Expected Value, along with some talk about Random Variable. In our example the random variable is the number written on the dice whose values can be x(1),x(2),…,x(n).

. You can use the above formula for any experiment, not just dice.

Let’s solve some problems with what we have learned so far.

*How to Make an Animation CSS Neon Love – CSS JavaScript Tutorial*

*How to Make an Animation CSS Neon Love – CSS JavaScript Tutorial*

Problem 1:

Suppose you have a coin, one side of this coin is also slightly heavier, the probability of getting heads is 0.4 if the coin is tossed once.

And the probability of getting a tail is 1−0.4=0.6

. You toss the coin until you get heads. How many times on average (expected) do you need to toss a coin to get heads?

Assume that on average E

If you toss a coin twice you will get heads. It should be remembered here that each coin toss is an independent event, that is, the result of one toss has no relation to the result of the next toss. Now we can have two results of the experiment:

You get a tail, lose a toss, you still average E

The bar must be tossed. Expected Value 0.4(1+E)

. The concept of recursion comes into play here.

You got a head, you don’t need to toss anymore. Expected Value (0.6)⋅(1)

.

On average you get E=0.4⋅(1+E)+(0.6)⋅(1)

Toss the coin twice. If we write it in reverse we get 1.6667. This means if you experiment an infinite number of times you will get heads after an average of 1.6667 tosses.

Problem 2:

You have a coin tossed with probability p(h) of getting heads or tails.

and p(t)

. How many times (expected) to toss to get two heads in a row?

Think E

If you toss times you will get two heads in a row. Now if you get tails for the first time then one toss is wasted and you have to toss p(t)⋅(1+E) more times on average. But if you get heads the first time two things can happen, next time you get tails and need to toss p(t)⋅(1+E) more times, or next time you get another heads and need not toss anymore. Adding all events together, the expected value is E=p(h)⋅(1+p(t)⋅(1+E)+p(h)⋅1)+p(t)⋅(1+E).

If the coin is a fair coin, i.e. p(h)=p(t)=0.5 then the value of E will be 6

, you can check.

Now the question is n consecutive instead of two

How many times do you have to toss if you want to get the head? Think about it, if not, there is an answer on this site.

Problem 3:

A coin n

How many heads do you expect to get when the bar is tossed?

First think if the coin is tossed 1 time how many heads will you get on average (expected)? 0.5 probability that you get 1 head, rest 0.5 probability that you get no head. Then the expected value is 0.5⋅1+0.5⋅0=0.5

. This means that if you do the experiment an infinite number of times and toss the coin 1 time per experiment, on average you get 0.5 each time.

Get the T head.

n

If you toss times, you need to add this value n times: (0.5⋅1+0.5⋅0)+(0.5⋅1+0.5⋅0)+…(n times)=n⋅0.5. This means that if you do the experiment an infinite number of times and toss the coin n times per experiment, on average you get n×0.5 each time.

Get the T head.

Another similar problem is, n

There are t students, each is asked to write a number between 1 and 100. If the experiment is carried out an infinite number of times, on average, how many students will write any number between 1 and 9? Assume that each number has an equal probability of being written (although this would not be true if the experiment were actually carried out, people prefer some numbers!).

Problem 4:

n

Expected number of coin tosses to get heads?

We will solve this problem recursively. Think n

To get a head, you have to toss the coin E(n) times. Now if I get a head then I need n−1 more coins for which I have to toss E(n−1) more times. But if I get a tail then I get more E(n).

The bar must be tossed.

Then the total number of coins to be tossed is E(n)=0.5⋅(1+E(n−1))+0.5⋅(1+E(n))

times Simplifying this gives E(n)=E(n−1)+2. Now we need a base case to stop the recursion. If we get 0 heads then there is no need to toss, E(0)=0

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