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 Contents: Watch the video for three examples: Probability: Dice Rolling Examples Watch this video on YouTube. Can’t see the video? Click here. Need help with a homework question? Check out our tutoring page! Dice roll probability: 6 Sided Dice ExampleIt’s very common to find questions about dice rolling in probability and statistics. You might be asked the probability of rolling a variety of results for a 6 Sided Dice: five and a seven, a double twelve, or a double-six. While you *could* technically use a formula or two (like a combinations formula), you really have to understand each number that goes into the formula; and that’s not always simple. By far the easiest (visual) way to solve these types of problems (ones that involve finding the probability of rolling a certain combination or set of numbers) is by writing out a sample space. Dice Roll Probability for 6 Sided Dice: Sample SpacesA sample space is just the set of all possible results. In simple terms, you have to figure out every possibility for what might happen. With dice rolling, your sample space is going to be every possible dice roll. Example question: What is the probability of rolling a 4 or 7 for two 6 sided dice? In order to know what the odds are of rolling a 4 or a 7 from a set of two dice, you first need to find out all the possible combinations. You could roll a double one [1][1], or a one and a two [1][2]. In fact, there are 36 possible combinations. Dice Rolling Probability: StepsStep 1: Write out your sample space (i.e. all of the possible results). For two dice, the 36 different possibilities are: [1][1], [1][2], [1][3], [1][4], [1][5], [1][6], [2][1], [2][2], [2][3], [2][4], [2][5], [2][6], [3][1], [3][2], [3][3], [3][4], [3][5], [3][6], [4][1], [4][2], [4][3], [4][4], [4][5], [4][6], [5][1], [5][2], [5][3], [5][4], [5][5], [5][6], [6][1], [6][2], [6][3], [6][4], [6][5], [6][6]. Step 2: Look at your sample space and find how many add up to 4 or 7 (because we’re looking for the probability of rolling one of those numbers). The rolls that add up to 4 or 7 are in bold: [1][1], [1][2], [1][3], [1][4], [1][5], [1][6], There are 9 possible combinations. Step 3: Take the answer from step 2, and divide it by the size of your total sample space from step 1. What I mean by the “size of your sample space” is just all of the possible combinations you listed. In this case, Step 1 had 36 possibilities, so: 9 / 36 = .25 You’re done! Two (6-sided) dice roll probability tableThe following table shows the probabilities for rolling a certain number with a two-dice roll. If you want the probabilities of rolling a set of numbers (e.g. a 4 and 7, or 5 and 6), add the probabilities from the table together. For example, if you wanted to know the probability of rolling a 4, or a 7:
Probability of rolling a certain number or less for two 6-sided dice.
Dice Roll Probability TablesContents: Probability of a certain number with a Single Die.
Probability of rolling a certain number or less with one die.
Probability of rolling less than certain number with one die.
Probability of rolling a certain number or more.
Probability of rolling more than a certain number (e.g. roll more than a 5).
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Visit out our statistics YouTube channel for hundreds of probability and statistics help videos! ReferencesDodge, Y. (2008). The Concise Encyclopedia of Statistics. Springer.
Need help with a homework or test question? With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. Your first 30 minutes with a Chegg tutor is free! Comments? Need to post a correction? Please Contact Us. Explanation: We have blue, red, and green marbles in a bag. We need to consider all of the ways that we could draw at least one red. This means we can draw a red the first time, a red the second, or a red both times. These are the possible ways we could at least draw one red marble: We can draw a red and then a blue. We can draw a red and then a green. We can draw a blue and then a red. We can draw a green and then a red. We could draw a red and then another red. So we need to find the probability of each of these five scenarios. Then, we need to add these probabilities. Let's look at the probability of the first scenario (red, then blue). The probability of drawing a red on the first time would be 4/10, because there are 10 marbles, and four are red. On the second draw, we don't put this marble back. This means we only have 9 marbles in the bag, and four of them are blue. Thus, the probability of the second draw being blue would be 4/9. The probability of drawing a red and then a blue is equal to the product of these two events. Whenever we want to find the probability of one event AND another, we need to multiply. Thus, the probability of drawing the red AND then a blue would be (4/10)(4/9) = 16/90. We can calculate the probability of the other four possibilities in a similar fashion. The probability of drawing a red and then a green is (4/10)(2/9) = 8/90 The probability of drawing a blue then a red is (4/10)(4/9) = 16/90 The probability of drawing a green then a red is (2/10)(4/9) = 8/90 The probabilty of drawing a red then a red is (4/10)(3/9) = 12/90 To find the total probabilty, we need to add up the probabilities of the five different scenarios. Whenever we want to find the probability of one event OR another, we add. So the final probablity is 16/90 + 8/90 + 16/90 + 8/90 + 12/90 = 60/90 = 2/3. The probability of drawing at least one red is 2/3. Page 2
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A coin is flipped 4 times. What is the probability of getting 4 heads in a row? Possible Answers:
1/16 1/100 1/2 None of the other answers 1/4
Correct answer: 1/16 Explanation: The probability a heads will show on any flip is 1/2. To find the probability of 4 consecutive heads, multiply the probability of each individual flip together. 1/2 * 1/2 * 1/2 * 1/2 = 1/16
A restaurant offers two different entrees, four different side dishes, and three desserts. If a customer wants to order a meal special that consists of one entree, two different side dishes, and one dessert, how many different meals are possible? Possible Answers:
Correct answer: 36 Explanation: Our strategy is as follows: We must determine the number of possible entrees, the number of possible combinations of two side dishes, and the number of desserts. Then we must multiply these four numbers together to get the number of all of the possible meals. Let's label the two different entrees E1 and E2 The customer can choose one of these, so he has two different options for his entree. Let's label the four different side dishes S1, S2, S3, and S4. The customer can only choose two of these four dishes, and the order in which he chooses doesn't matter. For example, if he were to choose mashed potatoes and broccoili, he would end up with the same meal as if he were to choose broccoli and then mashed potatoes. So we need to consider all of the different pairs that can be selected from these four side dishes. Here are all of the possible combinations: S1 and S2 S1 and S3 S1 and S4 S2 and S3 S2 and S4 S3 and S4 This means that the customer has six different combinations of two side dishes. Finally, the customer can choose one dessert from three different choices. This means there are three possibilities for dessert. So, the customer has two different options for entrees, six different options for the two side dishes, and three different choices for dessert. To find the total number of meals possible, we can multiply these three numbers together. Number of meals = 2 x 6 x 3 = 36. The answer is 36.
How many positive four-digit integers have the thousands digit equal to 8 and the units digit (ones digit) equal to 5? Possible Answers:
Explanation: The number of possibilities in a joint event (that is, the entire four-digit integer) is the product of the number of possibilities of each individual event (the number of possibilities for each digit individually). If WXYZ represents a 4-digit integer, the problem tells us that there is only one possible value for W (which is 8) and only possible value for Z (which is 5). We also know that there are ten possibilities each for X and Y (they can be the digits 0-9). So the number of possible values for WXYZ is the product of the possible values for each digit, which is 1 * 10 * 10 * 1 = 100.
Nora is packing for a trip. Of the scarves in her closet, 8 are blue. She will randomly pick one of the scarves to pack. If the probability is 2/5 that the scarf she will pick is blue, how many scarves are in her closet? Possible Answers:
Explanation: We are told that there are 8 blue scarves in the closet, and that the probability of a blue scarf being chosen is 2/5. So, if T represents the total scarves in the closet, we know that 8/T = 2/5. Cross-multiplying gives: 40 = 2T And dividing both sides by 2 gives: 20 = T Therefore, Nora has a total of 20 scarves in her closet.
Two dice are rolled. What is the probability that the product of the numbers rolled is 15? Possible Answers:
Explanation: The only possibility to roll a product of 15 is to roll a 5 on the first dice and a 3 on the other, or a 3 on the first and 5 on the second. There are 36 total possibilities for two dice (6 * 6), 2 possibilities out of 36 gives you 2/36 = 1/18.
A jar contains three red marbles, four blue marbles, and six green marbles. Jennifer draws a marble and then draws a second one without replacement. What is the probability that she will draw a blue marble and then a green marble? Possible Answers:
Explanation: There are three red marbles, four blue marbles, and six green marbles. This means that there are a total of thirteen marbles. The probability of drawing a blue marble on the first drawy would be 4/13. Because Jennifer doesn't replace the marble, after she draws the blue marble, there are only twelve marbles left. This means that the probability of next drawing a green marble would be 6/12 = 1/2. To find the probability of the two events happening together, we must multiply them. In general, when you want to find the probability of one event AND another, you must multiply. probability = (4/13)(1/2) = 2/13 The answer is 2/13.
The Sugar Shak has a "make your own sundae" bar. You can choose one of three ice creams (strawberry, chocolate, or vanilla), one of three sauces (strawberry, carmel, or chocolate), and one of four toppings (peanuts, whipped cream, cherry, or M&Ms). How many different sundaes can be made? Possible Answers:
Explanation: We have three choices for ice cream, three choices for sauces, and four choices for toppings. Each selection is an independent event, so the choices are multiplied together: 3 * 3 * 4 = 36.
A bag contains two white marbles, five green marbles, and three red marbles. What is the probability of picking two red marbles if replacement is not allowed? Possible Answers:
Explanation: Probability is a number between 0 (will not happen) and 1 (will definitely happen). Probability is generally a fraction where the numerator is the total of what you want to happen and the denominator is the total count. Total ways to choose two red marbles: 3 * 2 = 6 Total ways to choose two marbles: 10 * 9 = 90 Therefore, the probability of choosing two red marbles is 6/90 or 1/15
What is the probability of choosing three hearts in three draws from a standard deck of playing cards, if replacement of cards is not allowed? Possible Answers:
Explanation: The standard deck of cards has 52 cards: 13 cards in 4 suits. Ways to choose three hearts: 13 * 12 * 11 = 1716 Ways to choose three cards: 52 * 51 * 50 = 132600 Probability is a number between 0 and 1 that is defines as the total ways of what you want ÷ by the total ways The resulting simplified fraction is 11/850
A lottery is being run at a high school to allocate parking spots. The school has 200 seniors, 300 juniors, 350 sophomores, and 450 freshmen. Each eligible senior will have their name entereted into the lottery twice, with all other eligible students' names being entered once. Only juniors and seniors will be eligible for parking spots. If there are 150 parking spots, what is the probability that any given junior will receive a spot? Possible Answers:
Explanation: Find the probabilty a junior's name will be pulled for a single lottery trial. Then calculate the probability given 150 lottery trials. (200 seniors * 2 entries each) + 300 juniors = 700 entries For any single junior then, the odds are 1/700 for a single lottery trial. For 150 trials, a junior will have (1/700 * 150 trials) = 150/700, which simplifies to 3/14
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When two dice are thrown, what is the probability that the sum of the numbers is a multiple of four?
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