Probability

an irregular variation is one which occurs predictably and at random, usually over a very short time period. it is impossible to allow for irregular variations in forecasting rather, they are used to explain abnormalities after the event.

hope this helps :-)

Expected number of heads is 1/4 * 32 or 8 heads.

Assuming you're drawing from a complete shuffled 54-card deck, chances are one in twenty-seven.

The odds of getting 3 heads in a toss of 3 coins is 1 in 8, or 0.125.

Each coin is probabalistically unrelated to each other, so you simply multiply the odds for each coin. 1 in 2 times 1 in 2 times 1 in 2 is 1 in 8.

I suppose you mean, at least one of those numbers. Just calculate the probability of NOT getting any of those, and take the complement. The probability of not getting a one nor a five on a single die is 4/6 or 2/3. For two dice, the probability is 2/3 x 2/3 = 4/9. So, the probability of getting at least a one or a five with two dice is 1 - 4/9 = 5/9.

I suppose you mean, at least one of those numbers. Just calculate the probability of NOT getting any of those, and take the complement. The probability of not getting a one nor a five on a single die is 4/6 or 2/3. For two dice, the probability is 2/3 x 2/3 = 4/9. So, the probability of getting at least a one or a five with two dice is 1 - 4/9 = 5/9.

I suppose you mean, at least one of those numbers. Just calculate the probability of NOT getting any of those, and take the complement. The probability of not getting a one nor a five on a single die is 4/6 or 2/3. For two dice, the probability is 2/3 x 2/3 = 4/9. So, the probability of getting at least a one or a five with two dice is 1 - 4/9 = 5/9.

I suppose you mean, at least one of those numbers. Just calculate the probability of NOT getting any of those, and take the complement. The probability of not getting a one nor a five on a single die is 4/6 or 2/3. For two dice, the probability is 2/3 x 2/3 = 4/9. So, the probability of getting at least a one or a five with two dice is 1 - 4/9 = 5/9.

i THINK IT IS .05

Every discipline of mathematics has contemporary problems and theories being worked on. Some examples of the more popular mathematics topics being worked on now are Lie algebras, knot theory, and game theory.

Seven.

216... there are 6 outcomes for each cube, so 6^3 is 216

There are 36 possible outcomes (6 X 6). the probability of getting a one is 1/6; the probability of getting a four is 1/6. 1/6 X 1/6 = 1/36. Then, you multiply 1/36 by 2 because you can either get a 4 and a 1 or a 1 and a 4. Therefore, the answer is:

2/36 = 1/18

Since there are 6 numbers (1-6) on each die, and you're rolling it 5 times, the total number of possibilities is 65. You now have to find out, out of the total, how many possibilities fill your requirements. For 5 rolls of one the chance is 1/65. There are 5 possible sets of rolls where you'll get 4 1s (4 1s and a 2, 3, 4, 5, or 6). Lastly there are 25 possible sets of rolls where you'll get 3 1s (3 1s and a 2 and 2 or a 2 and 3 or a 2 and 4 etc). so adding up all the possibilities that is 31/65. That is a percentage of .4% or 4 out of 1000 rolls of all the dice you will get 3, 4, or 5 1s. Sounds like a yahtzee question.

4-6 Chances

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From Rafaelrz.

There are three even numbers in a regular die, 2, 4 and 6.

There are six equally probable outcomes of which three are

unfavourable.

So the odds of rolling an even number are 3:3 or 1:1.

This means there is a 50% probability of success.

Let your raw score be x and M the mean and S the standard deviation.

The Z score for your specific x is Z=(x-M)/S

So say your score is 80 (out of 100) and the mean is 70 and the standard deviation is 10.

Then the z score for your 80 is:

(80-70)/10=1

If on the other hand you got a 60, then the z score would be -1.

As n increases the sampling distribution of pˆ (p hat) becomes approximately normal.

Yes.

It happened after dice had been used in game playing for almost 3,000 years. Sometime around

that epoch, or close to it, the dice manufacturers, at their annual convention, had to face the

situation that they had all ignored and repressed for all those centuries ... the game was just

not popular any more, and their sales of dice ... never a market leader ... were now totally sinking

through the floor. I mean they were all taking the pipe. Where originally there had been a dice

manufacturer for every two or three caves, now there were whole cities without a single one.

In spite of repeated government stimulus, bailout, welfare, and designation as historic structures,

these factories had steadily closed in droves throughout the length and breadth of recorded

human history, and they continued to do so. It wasn't only a matter of financial survival either ...

if you could see the kind of mindless drivel that had replaced the game of dice, you'd understand

that it was just as much a matter of honor, of manhood, of self-esteem. Wherever you looked,

whole families were turning away from their dice, getting more excitement and tittilation out of

burying stones, chewing their toenails, and throwing apples in the river. A pall hung over

that year's dice manufacturers' convention, like a pall. They knew that someone had to do

something in a hurry, or this convention would be their last; next year at convention time,

they'd have to spend the week at the office, or ... even worse ... stay home.

It was against this backdrop of seeming hopelessness that Murray ... a newcomer to the industry ...

mounted his campaign for the presidency of the dice manufacturers' trade association, there at the

annual convention. It didn't take much for Murray to run away with the election. He had the vote

locked up the first time he took the podium, and pointed out to everyone that even though there was

still plenty of crap, they weren't even doing it with dice at their own convention ! Murray's solution

to the whole problem was a simple one, as some of the best and most revolutionary ideas in history

have often been. Instead of the classic six-sided, symmetrical, featureless, polished cubes that

they had been manufacturing and peddling for centuries, Murray explained how a total, cataclysmic

turnaround in the dice market would follow a simple modification to the product ... assigning a

number from 1 to 6 to each face, and carving the corresponding number of shallow depressions

into the face.

My tale has already exceeded any amount of the dear reader's time and attention that I could

reasonably have requested, and so it must end without further details. It remains only to mention

that Murray obviously had it right. Dice became what they is today. The game, the game piece, and

the production lines that continue to bring it to us, were all revived. And if there remain any who

would doubt the value of Murray's contribution to the modern game, they need only consider how it

might be different if dice still had no spots. See what I mean ?

58=390625

There are five choices for each of the questions. To find the total number of ways to answer multiply 5 for each of the eight questions.

Formula for standard error (SEM) is standard deviation divided by the square root of the sample size, or s/sqrt(n). SEM = 100/sqrt25 = 100/5 = 20.

The ratio of girls to total students is 15:25, or 3:5. Three out of five students are girls so there would be a 60% probability that a girl would be chosen; a 2 out of 5 chance, or 40% probability that a boy would be chosen.

No, it can't.

If you roll two dice and add the numbers together, you might get 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, so there are eleven possibilities total. One is not included in this list because that would mean the first dice would roll a one and the second dice would roll a...zero? That's impossible!

7/128, or about 5.5%

The student has a 1/2 probability of getting each question correct. The probability that he passes is the probability that he gets 10 correct+probability that he gets 9 correct+probability that he gets 8 correct:

P(passes)=P(10 right)+P(9 right)+P(8 right)=[(1/2)^10]+[(1/2)^10]*10+[(1/2)^10]*Combinations(10,2)=[(1/2)^10](1+10+45)=56/1024=7/128.

A brown dwarf.

A brown dwarf.

A brown dwarf.

A brown dwarf.