Probability

its easy idiots, you use it when guessing :)

Probability is used to extrapolate the likelihood of a future event, so if you think about it, it's used all the time by everyone everyday.

Here are some main applications:

- Data Analysis

- Quantum Mechanics

- Thermodynamics

- Meteorology

- Social Science

- Business ans Finance.

The list is near endless.

The function of ileum is absorb vitamin B12 and bile salts. If products were not absorbed by the jejunum the ileum would absorb them.

From the related link, you can read directly the probability that Z is less than 1.51 is 0.9345.

1 in 36

Probability not a club = 1 - probability it is a club = 1 - 13/52 = 1 - 1/4 = 3/4.

Random lottery numbers, every time! I think it's great to mix it up, but I also like to see my numbers before I buy my ticket. I play the euro millions a bit more than the lottery as the rollover jackpots always look appealing! In theory it doesn't matter how you select the numbers so long as you pick the rquired number of numbers.

15..... 1,1,5 1,2,4 1,3,3 1,4,2 1,5,1 2,1,4 2,2,3 2,3,2 2,4,1 3,1,3 3,2,2 3,3,1 4,1,2 4,2,1 5,1,1

Br-. Bromine will grab an electron to make itself more stable, which makes it a negative ion.

40.

If you picked 3 aces, 3 two, 3 threes....all the way to 3 kings you would have 39 cards. The fourtieth card must make a four of a kind.

On a roll of two die, there are 6*6 = 36 possible outcomes. You can write each combination down on a piece of paper in a format such as (roll die 1, roll of die 2). In the probability section of the related web site, it lists the possible values (sum of the two die) and the probability. The probability is calculated on the number of different ways to get the sum listed. For example, there is 1 way to get a 2: (1,1). So the probability of rolling a 2 is: 1/36. The probability of rolling a 7 is: (6,1) & (5,2) & (4,3) & (3,4) & (2,5) & (1,6) = 6 ways to get a seven divided by 36 = 6/36 = 1/6. The mean is defined by the sum of each possible outcome times its probability. The formula in symbols is: μ = Σ x P(x) or: 2(1/36) + 3(2/36) + 4(3/36) + 5(4/36) + 6(5/36) + 7(6/36) + 8(5/36) + 9(4/36) + 10 (3/36) + 11(2/36) + 12(1/36) = 252/36 = 7 = mean.

Do you mean what are all the possible outcomes? Or what is the probability of a certain outcome? Need a little more information.

50 in 52, or 25 in 26.

Leonhard Euler ,Gaspard Monge ,Jean-Victor Poncelet , Jakob Steiner

Assuming that you cannot repeat any of the numbers that you are using to guess (e.g., would not guess 1, 4 and 4) AND that a match only counts once (e.g., in the serial number A23345678D, the matching 3 only counts as a single match despite appearing more than one time), the probability is between 15.4% and 15.5% (via 5MM simulations).

If we loosen the rules and say that there can be double matches with the numbers that you are guessing (e.g., in the serial number above, if 3 was a guess number, there would be two matches, not just one), the probability is between 25.4% and 25.5% (via 5MM simulations)

The spirit of the question suggests that we would never repeat a guessed number.

This is a very famous problem in probability and the answer goes against the intuition of most people including mathematicians. So picture 3 doors and a prize behind 1 of the doors. Let's say it is the WikiAnswers gold badge, worth lots of time and work Now since there are 3 doors and 1 prize, the chance of picking the gold badge is 1/3 or 33%. So I could predict that one time out of three, you pick the door with the prize. Now here is where it gets fun. In any probability problem, including this, the sum of the probabilities must equal 1 or 100%. If we pick one door the chance of finding the gold badge, is 1/3 as we said. If we look at all 3 doors, the chance of finding the prize is 1 or 100% since we know the badge is behind one of them and if we look at 2 doors, the chance of finding the prize is 2/3. So now you pick a door. There are only two choices. Either you picked the winning door or you did not. No other choices! Now Mathdoc, ok it was really Monty Hall, opens up one of the doors and reveals that there is no prize there. Monty has not just eliminated one of the three doors, he has eliminated one of the doors with a prize! This detail is important. Now say your initial choice was correct, you picked the door with the gold WikiAnswers badge. Now since you picked the right one, the other two are wrong ones. Now Monty opens one of them, does not really matter which. So if you stay with your first choice you are right 1/3 of the time. Now, let's say you picked wrong to start with. The door you pick does NOT have the gold badge. That means 2/3 times your initial choice is wrong as we said above. We also explained above that if we look at 2 doors, 2/3 times we can pick the prize. So now we have the prize behind one of those 2 doors. So since you picked the wrong door to start with this means that 2/3 of the time the prize is behind one of other two doors. So looking at those two doors that you did not pick as one probability unit, there is a 2/3 chance of picking the prize, but Monty is kind and has shown you where it is not which means it must be behind the other door. Remember there are only two doors you did not pick and one of the two has the gold and Monty showed you the one that does not so the other one does! Now think of the remaining door as inheriting the probability that the two doors had. That is to say, the remaining door now has a 2/3 probability of having the prize. Your initial chance of being right is 1/3 and the door remaining has 2/3 chance of being right.. Should you switch? Hmm I think so! So why is it not 50% you asked? That is a good question and important to understand in order to grasp the problem. The idea is that Monty shows you a losing door, so one door is right and one is wrong; so there should be a 50 % chance of picking the correct door. It seems that it does not matter if you switch or not. Here is why the intuition fails. Monty always has to open a losing door. So one losing door is always eliminated. Because this is a probability problem, the probability of your initial choice being correct and your remaining choices must equal 1 or 100%. That means if your initial choice being correct probability is 1/3, the probability of the other two being correct is 2/3 ( 1-1/3) If a door is not opened then there are two choices to switch to, it is kind of like just changing your mind. But this is not what happens. If it did you would just multiply 1/2x 2/3. Since Monty opens one door, you only have 1 to pick from, so instead of your odds being 1/2 you now need to multiply by 1. (you can pick form two doors, 1/2, you pick from one door 1 choice) So you must multiply 2/3 x1 NOT 1/2. That gives us a probability of 2/3 or 66% So it is always better to switch! To review in the more classic Monty method. The car is behind door C, so A and B have a goat. You get a car if you win and a goat if you lose. Say you pick the door with the car. You are then shown either door A or B with the goat. If you change you lose, if you stay you win. Now say you pick door A. Now I show you door B with a goat. If you switch you win, if you stay you lose. Third possible scenario, you pick door B and you are shown the goat behind door A. Same thing, if you change doors you win if you don't you lose. Of course each of the three options above has a 1/3 chance of happening. Since you are equally likely of picking any of the three. Now in the above scenarios, you win 2/3 times if you change. In 1/3 scenarios you win if you don't change. So when you switch 2/3 times you win the car that is 2/3 or 66%. YES THIS IS NOT OBVIOUS, that is why so many people became goat herders on the show.

50/50. There are two sides (heads and tails), so half of the time it will land on heads. 49.5% or something like that because the coin can land on heads, tails, or on its edge. but the likelihood is like a fraction of a percent, but it is possible

If the card is drawn randomly, the probability is 1/4.

Assuming you have got a normal deck with 2 Jokers (52+2 cards). Identify the probabiities of each event: Move 1. 1st Ace: 4/54 move 2. 2nd Ace: 3/53 Move 3. 3rd Ace: 2/52 Move 4: 4th Ace: 1/51 Move 5: Joker: 2/50 Now, u just have to get a product of all the above outcomes' probabilities. The answer u get is 1/7906275 = 1.26481813496242E-07

I guess it depends on whether you mean the actual amount of diamonds on each card, or just how many cards in the suit. A traditional deck of cards has A-K in each suit, which is 13 cards... 13 diamonds, and 13 each of the other suits as well.

If you mean how many actual diamonds on the cards, that depends on the artwork, but I'm guessing 55 for the amounts themselves (one through 10, since J-Q-K don't have the number of diamonds on them), plus 2 each for the little ones in the corners to show the suit (26), so approximately 81. Could be more, or less, depending on the individual artwork.

The possible ways to get less than 5 are:

1 + 1, 1 + 2, 1 + 3, 2 + 1, 2 + 2 or 3 + 1. So there are 6 ways, there are 36 possible results so the probability is 6:36 (or 1:6).

The probability of any event lies in the interval [0, 1].

The sum (or integral) over all possible outcomes is 1.

i did this research and made my own equation on my own, take our Galaxy for example

1 galaxy=150-350 billion stars

lets take 250 billion stars, and if each star has a solar system of 1 to 9 planets, lets say each star has 5.

5 planets times 250 billion stars = 1,250 billion planets

and if at least 1 in 5 planets has life on it

1/5 X 1,250 billion planets = 250 billion planets

so i think maybe at least 250 billion planets in OUR galaxy have life

but that's only our Galaxy (Milky Way) and in the universe there are billions of galaxies maybe trillions, and how do we know this is the only universe, could there be more?

(i) P(X <= 2, Y = 1) = P(X=0, Y=1) + P(X=1, Y=1) + P(X=2, Y=1)

= (0+1)/30 + (1+1)/30 + (2+1)/30 = 6/30 = 1/5.

(ii) P(X + Y = 4) = P(X=2, Y=2) + P(X=3, Y=1)

= (2+2)/30 + (3+1)/30 = 8/30 = 4/15.