What are the considerations for the choice of sample size?
Some considerations are:
13
If two fair dice were rolled what is the probability of them rolling a sum of 11?
x
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
Therfore, there is a 2 in 36 (or 1 in 18) chance of rolling a total of 11.
When tossing 2 fair 6 sided number cubes what is the probability of tossing a sum of ten?
The first cube can land in any one of 6 different ways.
The second cube can land in any one of 6 different ways.
Together, they can land in any one of (6 x 6) = 36 different ways.
Sum of ten can come up in 3 different ways:
4 - 6
6 - 4
5 - 5
The probability of tossing a ten is (3/36) = 1/12 = 8-1/3 %
In a bag there 2 red 3yellow 4green and 6 blue marbles What is the probability of P Blue?
6 out of 15
When determining the probability that two events happen at the same time, you convert the percents to decimals and then multiply the percents together. Therefore, 30 percent, or .3, times 50 percent, or .5
.3 x .5 = .15
Converting back into a percentage, the answer is 15% probability that you will get both. 10% is therefore incorrect.
Can parents be B pos and have a kid A B p?
No. If both parents are B+ then their children will be B+ (75%) or O+ (25%).
Would it be possible for one to use a gigabyte in one months time with a mobile laptop?
It depends on what you are doing. If you download a lot of movies or music or take a lot of pictures, you can use a gigabyte in a few hours or a day. If you do nothing but type up word documents on your computer, you might not have over a gigabyte of typed documents in a year's time. Yes, it is possible to use a Gigabyte a month, it is also possible to use a gigabyte an hour or a gigabyte a year, it just depends on what you do on your computer and how you are using it.
What is the formula for geometric probability?
If you have an experiment in which the probability of success at each trial is p, then the probability that the first success occurs on the nth trial is
Pr(N = n) = [(1 - p)^(n-1)]*p for n = 1, 2, 3, ...
How do you get the probability of 0?
The simplest way to get a probability of zero is to have an impossible event. For example, with a die (that's the singular of "dice") that has the numbers 1-6, getting each of these numbers has a probability of 1/6; getting the number has a probability of 0.There is another way to get a probability of zero, which occurs for certain situations that involve infinity; if you are prepared to read through some somewhat intense math, check the Wikipedia article on "Almost surely" for more details. (In this case, "almost surely" means a probability of 1; while "almost never" means a probability of 0.)
If three number cubes are rolled all at once how many ways can you get 2 of the same number?
I'm going to assume you mean standard dice, all three distinguishable, and exactly two of the same number.
First break the problem into smaller parts. First, determine the odds of getting the same number on the first two dice, and a different number on the third. Then just multiply this result by 3 and 216 to account for the first and last dice being the same, or the last two dice, then compute the total number of possibilities.
P(two dice being the same and last being different) = P(second die being the same as the first) * P(third die being different than the other two) = (1/6)(5/6) = 5/36
3*216*5/36 = 90 possibilities out of 216
What are the odds of getting a number graeater than 4 on a dice?
The chances of getting a number greater than 4 on a standard 6 face dice is a 2/6 chance which simplifies to a 1/3 chance. This is because their are only 2 numbers greater than 4.
If a random variable X has a Normal distribution with mean M and standard deviation S, then
Z = (X - M)/S
What is the probability of getting a sum of 7 and 11 when two number cubes are rolled?
Pr(7) = 1/36
Pr(11) = 2/36 = 1/18
What is the probabilty of a number lower than 25 being odd or a factor of 4?
The question is being answered assuming that by "number" you mean a positive integer. The probability that the number meets these requirements is 14/24 or 7/12.
What is the total number of spots on a six- sided regular die?
This is your lucky day ! I'm about to not only hand you the answer, but I've decided
to take my time and lay out the entire calculation for you as well, too:
1 + 2 + 3 + 4 + 5 + 6 = 21
Find the number of times the face 5 is expected to occur in a sequence of 2000 throws of a fair die?
The probability of 5 is 1/6
So the expected number of times the face 5 shows up in 2000 throws is 2000*1/6 = 333.3... .
The first thing to do is find how many possible combinations of those numbers match the criteria. All numbers that start with 7 match it. And there are 5x4x3=60 possible numbers that start with 7. All numbers that start with 5 also match it, and there are 5x4x3=60 numbers that start with 5. All numbers that start with 35, 37 or 39 match it. And there are 4x3x3=36 numbers that do this. And finally, all numbers that start with 325, 327 and 329 match it, and there are 3x3=9 numbers that do this. In total this is 60+60+36+9=165 possible 4 digit numbers that match the criteria. The number of possible four digit numbers in total is 6x5x4x3 = 360.
This means the probability of a 4 digit number using only 1,2,3,5,7 and 9 being greater than 3251 but less than 8825 is 165/360 which can be simplified to 11/24.