What does estimated experimental probability mean please someone answer?
One way of finding the probability is to carry out an experiment repeatedly. Then the estimated experimental probability is the proportion of the total number of repeated trials in which the desired outcome occurs.
Suppose, for example you have a loaded die and want to find the probability of rolling a six. You roll it again and again keeping a count of the total number of rolls (n) and the number of rolls which resulted in a six, x. The estimated experimental probability of rolling a six is x/n.
What is the value of an invector-plus citiri special steel 30 barrel 3-12 receiver?
Never heard of a citiri.
A sample is a small part of anything or one of a number, intended to show the quality, style, or nature of the whole; specimen.
So, for instance, when you go to an ice cream shop and they give you a taste spoon with one flavor you want to try, that is a sample. It is a small portion designed to help you decide if you want more of it.
What is the probability of dying in a lake?
The truth is that alot of things come into play, how good of a swimmer are you, how is the weather, are you familiar with the lake and a host of other things but because this is kind of a binomial distribution of probability that means it has two outcomes, you drowned or you do not. So the chance is, statistically speaking, 50-50
What number sentence shows the probability of not rolling an even number on a cube?
idk i'm only in 5th grade my guess would be 3
What is the percentage of not getting the sum of 3 on two dice?
Of the 36 possible outcomes of a throw of two dice, all but two result in a sum other than 3.
So Prob(Not a three) = 100*34/36 = 94.44... %
If you Roll a 4 or higher on a six-sided number cube what is the probablity?
The probability of rolling a 4 or higher on a standard die is 1 in 2.
I could not find a good derivation of the mgf for the normal on the web. However, I found a complete derivation on page 270 of Probability and Statistics by DeGroot, 2002 Edition. Most university libraries should have a copy.
The first equation on page 270 goes from exp(a)exp(b) to exp(a+b).
The second equation is a direct result of "completing the squares" method. See page 281.
In third equation on page 270 DeGroot goes from exp(a+b) to exp(a)exp(b) which allows to define a constant exp(mu*t +1/2*sigma^2 * t^2) which he can pull out from the integral.
The "disappearing" of the square-root with 2pi occurs because the C as defined in DeGroot iin the fourth equation is the complete integral of the cdf of a normal distribution, for any value of t.
DeGroot states we can replace mu with mu+sigma^2*t, but I would have preferred
Let mu* = mu +sigma^2*t, then for any t, the cdf of the normal, N(mu*, sigma), integrated from minus infinity to positive infinity must equal 1.
Thus what appeared to be an integration requires no integration at all. I could restate the derivation as:
mgf(t) = f(t) * integral of N(x, mu*,sigma) from minus infinity to positive.
mgf(t) = f(t) * 1
mgf(t) = exp(mu*t +1/2*sigma^2 * t^2)
If somebody rolls 2 standard six-sided dice what is the probability he obtains a sum of 7 or 11?
2 out of 12
Ans:
1 2 3 4 5 6 1 2 3 4 5 6
total no N of sums of two numbers assuming dice are indistinguishable:
N = 36
ways to get 7:
1 and 6 with a chance of (1/6)*(1/6) = 1/36
2 and 5 wih a chance of 1/36
4 and 3 with a chance of 1/36
whats the chance of getting 7 with either of these combinations:
1/36 + 1/36+ 1/36 = 1/12
what about 11:
6 and 5, which is the only way to get an 11:
the prob of this is (1/6)*(1/6) = 1/36
the prob of getting a 7 or an 11, with two dice:
1/36 + 1/12 = 1/9!
Hopefully someone will jump in if I've made a mistake.
OK, I've been told the answer is 2/9. If you do the same working as above, but you'll have to assume there are two ways to get each sum:
example, if you get a 7 from 2 and 5, then you can also get 7 from a 5 and 2. This means the dice are distinguishable!
What is the probability of flipping a coin twenty times and only getting heads twice?
20!/(18!*2!) * (1/2)^20 = 190/1048576 = 0.000181198...
So less than 1 in 5000.
How do you get probabilities of situations that contain multiple events?
The answer depends on if and how the events depend on one another.
1+1 lowest, 6+6 highest so 12 - 2 = 10
Suppose there are only 20 cars in the world: 5 will be black and 2 will be black SUVs, so if you pick a black car at random there are 2 chances out of 5 that it will be an SUV, ie probability of 0.4 or 40%
False.
2/5
There are 20 numbers from 20 through 39, and 4 of them are prime (23, 29, 31, 37), the probability is 4 in 20 or 0.20.
What is the probability of throwing a total of four when you throw a pair of standard fair dice?
3 out of 36 possible rolls (1 & 3, 2 & 2, 3 & 1), so 3/36, or 1 out of 12.