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If two dice are thrown what is the probability that some of digits is ten?
Wiki User
∙ 9y ago
None of the digits can be 10, so the probability is 0.
Wiki User
∙ 9y ago
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How are dice thrown in games?
Dice can be thrown in games many different ways. Some games provide a dice cup that allows players to shake the dice and then toss them onto the table.
What are some probability games?
Black Jack, Poker, any dice game, Probhex, however it can also be educational than other probability games.
What is the probability of rolling a number greater than 18?
The answer depends on what you are rolling: three or more ordinary dice, or fewer dice with non-standard numbers on them, or a die with some other shape.
What are some probability questions for a nine sided dice?
Q1: You can have a die and many dice, but what is "a dice"? Q2: A 9-sided die cannot be a regular polyhedron. With an irregular polyhedron some outcomes are more likely than others. What is the shape? Q3: Since there are 2606 different (non-isomorphic) 9-sided dice possible and each one will have a different probability distribution, how will you determine the probability distribution function for the faces on your die? Q4: What are the numbers on the faces? Are they 1-9 or some other set, such as for a doubling die in backgammon? Once you have got your answers to the above you can start thinking about the probability distribution function for the different numbers on the die. Then you can look at questions regarding multiple rolls.
How do you do probability of independent events?
The answer depends on what you mean by "do". Does it mean calculate individually, calculate the probability of either one or the other (or both), calculate the probability of both, calculate some function of both (for example the sum of two dice being rolled)?
What is the probability of rolling a 7 with two dice?
The probability or rolling a 7 with one roll of a pair (2) dice is 0.166666.... We get this number by dividing 6 by 36 (or 1 by 6). Let's review what a probability is and then find out how we get the 6 and the 36.We know that probabilities range from zero (0) to one (1), inclusive. A probability of 0 means that a given result cannot happen. Like rolling a 13 with one toss of a pair of dice. The largest number that can appear is 12, so rolling a 13 cannot be done. The probability of that happening is 0.A probability of 1 means it must happen. Like rolling a number between 2 and 12, inclusive, with one roll of the dice. Rolling a 2, 3, 4, 5, 6 7, 8, 9, 10, 11, or 12, is something that must happen, so the probability of that is 1. All probabilities that something will happen will fall on or between 0 and 1. All of them. Some folks often interchange the term probability with "the odds" of something happening, and this might be the cause of some confusion. Now that we've reviewed probability, let's look at odds and see where the numbers come from.There are 6 ways to make a 7 with two dice. They are 1-6, 2-5, 3-4, 4-3, 5-2, and 6-1. And there are 36 (6 x 6) possible outcomes with a single roll of two dice. The odds of a 7 coming up are 6 in 36, or 1 in 6. As this is true, the probability of the 7 appearing will be the number of times the combination is possible (6) divided by the total possible combinations (36), which leads us to 6 divided by 36 or 1 divided by 6, or 0.166666....Use the link to the Related question to see about the odds of rolling that 7 with two dice.
What is the Role of Independence in the topic of probabilities and is there an example of it?
There is a wonderful and brief explanation at the link. One thought: Without the concept of independence, the accurate probability that two events will occur together would be a problem. You need to know if the events are dependent on one another in some way. If I roll two fair dice, what is the probability that I will roll two sixes? I know that the events (the results I get from the two dice) are independent of one another. So the probability of their happening together is the product of the probabilities that they will happen independently.
What are the Odds of rolling doubles?
Answer 1:The odds are very easy to calculate. Simply divide the number of "valid" rolls against all possible rolls. For ease, you can write down all possible combination for the 2 dice.1-1; 1-2; 1-3; 1-4....and so on, remember 1-4 and 4-1 are different rollsThere are 36 unique possible combination, and 6 of them are doubles, so that's 6/36 chances (and since 6 goes into 36, 6 times, this reduces to 1/6) or about 17%Answer 2:Another way to look at this problem, generically, is to assume we have an 'n' face dice. In most cases, dice have 6 faces (1, 2, 3, 4, 5, 6). But why not create a solution that works for any number of sides? Well, if we are trying to calculate the probability of rolling two dice (dice-1 and dice-2) of 'n' sides at the same time and having them turn up as doubles, only one of the dice really matters. Here's why. Dice-1 is guaranteed to land on a number 1-n. This will happen every time (on a fair dice, disregarding freak incidents). What we are trying to calculate is the probability that dice-2 will land on the SAME number as dice-1. Dice-2 can only land on one of 'n' values: 1, 2, 3, 4, 5, 6, ... , 'n'. For you non math folks, this just means it must land on a number from 1 to 'n' where 'n' is the number of sides on your dice. Out of all of the sides that dice-2 can PHYSICALLY land on, one of the sides MUST necessarily have the same as the value that dice-1 landed on. That is to say, if dice-1 landed on the value 3, there must be some chance that dice-2 will also land on the value 3. The probability of this occurring on a fair die is 1 divided by the total number of possible outcomes, which would be 'n'. So, really, there is a 1/n chance that dice-2 will land on the same number as dice-1. Thus, our probability for rolling doubles is simple 1/n. For our 6 sided dice example, our dice-1 lands on some value between 1 and 6 and there is a 1/6 chance dice-2 will match it.
What are the odds of rolling 6 same numbers in a six sided die?
The probability of rolling the same number six times on a standard die is (1 in 6)5 or 1 in 7776, or about 0.0001286. The reason the exponent above is five instead of six is that the probability of rolling "some" number on one die is 1, so you need to look at the probability of the other five dice matching the first die. It would not matter if you rolled one die six times, or six dice one time. The odds are the same.
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.)
What is the weight of a dice in grams?
One cube is a die, many are dice. So you cannot have one dice. The answer depends on what the die is made of and on how large it is. You can get sponge dice for rear-view mirrors in cars (why??!) which are lighter that some playing dice.
What is the experimental probability of rolling two dice 100 times and getting a sum divisible by three?
The successfule outcomes that fit your problem are a 3,6,9 and 12. There are 36 combinations, of which 3 can be the outcome of 2 events (1,2) and (2,1), 6 can be the outcome of 5 events, (1,5), (2,4), (3,3), (4,2), (5,1), 9 can be the outcome of 4 events (3,6),(4,5),(5,4), (6,3) and 12 is the outcome of 1 event (6,6). So out of the 36 combinations, we have 2+5+4+1 or 12 events, so 12/36 = 0.33. Now, if you throw two dice 100 times, and on the average this experiment should have 33.33 successes. Of course, some times you might have 41 successes, some times 35, but the long term average of 100 throws is 33.33 successes. This I would call the expected number of occurrences, not experimental probability. -- A bit extra to my answer: In experiments that involve chance, the results are never known. I might throw the dice 100 times and calculate a 40% of the time have success or a 25% of the time have success. These estimates are called proportions and are, I think, your "experimental probabilities" or sample estimates of success probability of your population. Of course, as given we know the success probability of the population (0.33). The binomial distribution can provide the probability of tossing two dice, "n" times and obtaining "x" successes, where the probability is 0.33. For example, I can state that 80% of the time, the number of successes will be between 27 and 39, when the dice are thrown 100 times using the binomial distribution. This is calculated by calculating the probability of 39 or fewer successes occur minus the probability of 27 or fewer sucesses occur. I have to use the cumulative distribution function (CDF). In Excel, I calculated: +binom(33-a1,100,0.33,TRUE)-binom(33-a1,100,0.33,TRUE) and varied a1 (whole numbers). When I tried a1=6, I obtained the 80%. At a1=8 (25 to 41 successes) I have the 91% confidence interval. The TRUE parameter means that I am using the CDF of the binomial distribution.
