Show
Solution: When two dice are thrown simultaneously, the sample space of the experiment is {(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)} So there are 36 equally likely outcomes. Possible number of outcomes = 36. (i)Let E be an event of getting a doublet. Favourable outcomes = {(1,1), (2,2),(3,3), (4,4), (5,5),(6,6)} Number of favourable outcomes = 6 P(E) = 6/36 = 1/6 Probability of getting a doublet is 1/6 . (ii)Let E be an event of getting a sum of 8. Favourable outcomes = {(2,6), (3,5), (4,4), (5,3), (6,2)} Number of favourable outcomes = 5 P(E) = 5/36 Probability of getting a sum of 8 is 5/36. > Suggest Corrections 16
Hint:Here the given question is based on the concept of probability. We have to find the probability of getting more than 7 when two dice are thrown at a single time. For this, first we need to find the total outcomes of two dice thrown at a time then by using the definition of probability and on further simplification we get the required probability of choosing a card. Complete step by step answer: Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e., how likely they are to happen, using it. Probability can range in from 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event. The probability formula is defined as the probability of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.$\text{Probability of event to happen}\,P\left( E \right) = \dfrac{\text{Number of favourable outcomes}}{\text{Total Number of outcomes}}$Consider the given question: The two dice are thrown at a single time then we have to find the probability of getting a more than 7. If the Three dice are thrown simultaneously$\eqalign{& \left\{ {\left( {1,1} \right)} \right.,\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),\left( {1,6} \right), \cr & \left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right), \cr & \left( {3,1} \right),\left( {3,2} \right),\left( {3,3} \right),\left( {3,4} \right),\left( {3,5} \right),\left( {3,6} \right), \cr & \left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right),\left( {4,6} \right), \cr & \left( {5,1} \right),\left( {5,2} \right),\left( {5,3} \right),\left( {5,4} \right),\left( {5,5} \right),\left( {5,6} \right), \cr & \left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left. {\left( {6,6} \right)} \right\} \cr} $Total number of outcomes$ = 36$A sum of 7 can be obtained in 6 ways =$\left\{ {\;(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} \right\}$A sum of 8 can be obtained in 5 ways $ = \left\{ {(2,6),(3,5),(4,4),(5,3),(6,2)} \right\}$A sum of 9 can be obtained in 4 ways $ = \left\{ {(3,6),(4,5),(5,4),(6,3)} \right\}$A sum of 10 can be obtained in 3 ways $ = \left\{ {\;(4,6),(5,5),(6,4)} \right\}$A sum of 11 can be obtained in 2 ways $ = \left\{ {\;(5,6),(6,5)} \right\}$A sum of 12 can be obtained in 1 way $ = \left\{ {\;(6,6)} \right\}$Therefore, a sum greater than 7 occurs when the total is 8,9,10,11 or 12. So, the number of possible outcomes to getting a more than 7 is: $5 + 4 + 3 + 2 + 1 = 15$ ways.By the definition of probability$ \Rightarrow \,\,P\left( \text{getting a score more than 7} \right) = \dfrac{\text{Total possible outcomes to get more than 7}}{\text{Total number of outcomes}}$$ \Rightarrow \,\,P\left(\text{getting a score more than 7} \right) = \dfrac{{15}}{{36}}$Divide Both numerator and denominator of RHS by 3, then we get$\therefore \,\,P\left(\text{getting a score more than 7} \right) = \dfrac{5}{{12}}$Hence, the required probability is $\dfrac{5}{{12}}$.Therefore, option C is the correct answer. Note:The probability is a number of possible values. Candidates must know the knowledge of dice, there are six faces in a single dice and if possible outcomes will be 6, then the total outcomes of two dice thrown simultaneously are $6 \times 6 = 36$. When we imagine the dice and its thrown the solution will be easy to solve.
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).
Back to top
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 post a comment on our Facebook page. |
In a single throw of two dice, what is the probability of getting a sum greater than 10
zusammenhängende Posts
Urheberrechte © © 2024 frojeostern Inc.
