Answer  Hint: To solve this question, we will start with finding the total outcomes of throwing dice three times, then we will find the favourable outcomes of throwing three dice. Afterwards using the formula, we will get the probability of getting the same number on the three dice. Complete step-by-step answer: We have been given, that we need to throw three dice. So, we need to find the probability of getting the same number on the three dice.We know that, \[Probability = \dfrac{{{\text{favourable outcomes}}}}{{{\text{total outcomes}}}}\]\[ \ldots .eq.\left( 1 \right)\]Now, outcomes of throwing single dice \[ = {\text{ }}\left\{ {1,2,3,4,5,6} \right\}{\text{ }} = {\text{ }}6\]So, total outcomes of throwing dice three times \[ = {\text{ }}6 \times 6 \times 6{\text{ }} = {\text{ }}216\]Now favourable outcomes of throwing three dice \[ = {\text{ }}\left\{ {1,1,1} \right\},{\text{ }}\left\{ {2,2,2} \right\},{\text{ }}\left\{ {3,3,3} \right\},{\text{ }}\left\{ {4,4,4} \right\},{\text{ }}\left\{ {5,5,5} \right\},{\text{ }}\left\{ {6,6,6} \right\}{\text{ }} = {\text{ }}6\]On putting the value of favourable outcomes and total outcomes in \[eq.\left( 1 \right),\]we getProbability of getting the same number on the three dice $ = \dfrac{6}{{216}}$$ = \dfrac{1}{{36}}$Thus, probability of getting the same number on the three dice is $\dfrac{1}{{36}}.$Note: In probability, the number of favourable outcomes is the total number of choices we are actually looking for and total outcomes are possible outcomes available.
Probability for rolling three dice with the six sided dots such as 1, 2, 3, 4, 5 and 6 dots in each (three) dies. When three dice are thrown simultaneously/randomly, thus number of event can be 63 = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its faces.Worked-out problems involving probability for rolling three dice: 1. Three dice are thrown together. Find the probability of: (i) getting a total of 5 (ii) getting a total of atmost 5 (iii) getting a total of at least 5. (iv) getting a total of 6. (v) getting a total of atmost 6. (vi) getting a total of at least 6. Solution: Three different dice are thrown at the same time. Therefore, total number of possible outcomes will be 63 = (6 × 6 × 6) = 216.(i) getting a total of 5: Number of events of getting a total of 5 = 6 i.e. (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1), (2, 1, 2) and (1, 2, 2) Therefore, probability of getting a total of 5 Number of favorable outcomesP(E1) = Total number of possible outcome = 6/216 = 1/36 (ii) getting a total of atmost 5: Number of events of getting a total of atmost 5 = 10 i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1) and (1, 2, 2). Therefore, probability of getting a total of atmost 5 Number of favorable outcomesP(E2) = Total number of possible outcome = 10/216 = 5/108 (iii) getting a total of at least 5: Number of events of getting a total of less than 5 = 4 i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1) and (2, 1, 1). Therefore, probability of getting a total of less than 5 Number of favorable outcomesP(E3) = Total number of possible outcome = 4/216 = 1/54 Therefore, probability of getting a total of at least 5 = 1 - P(getting a total of less than 5) = 1 - 1/54 = (54 - 1)/54 = 53/54 (iv)
getting a total of 6: Number of events of getting a total of 6 = 10 i.e. (1, 1, 4), (1, 4, 1), (4, 1, 1), (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1) and (2, 2, 2). Therefore, probability of getting a total of 6 Number of favorable outcomesP(E4) = Total number of possible outcome = 10/216 = 5/108 (v) getting a total of atmost 6: Number of events of getting a total of atmost 6 = 20 i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1), (1, 2, 2), (1, 1, 4), (1, 4, 1), (4, 1, 1), (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1) and (2, 2, 2). Therefore, probability of getting a total of atmost 6 Number of favorable outcomesP(E5) = Total number of possible outcome = 20/216 = 5/54 (vi) getting a total of at least 6: Number of events of getting a total of less than 6 (event of getting a total of 3, 4 or 5) = 10 i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1) (1, 1, 3), (1, 3, 1), (3, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1). Therefore, probability of getting a total of less than 6 Number of favorable outcomesP(E6) = Total number of possible outcome = 10/216 = 5/108 Therefore, probability of getting a total of at least 6 = 1 - P(getting a total of less than 6) = 1 - 5/108 = (108 - 5)/108 = 103/108 These examples will help us to solve different types of problems based on probability for rolling three dice. Probability Probability Random Experiments Experimental Probability Events in Probability Empirical Probability Coin Toss Probability Probability of Tossing Two Coins Probability of Tossing Three Coins Complimentary Events Mutually Exclusive Events Mutually Non-Exclusive Events Conditional Probability Theoretical Probability Odds and Probability Playing Cards Probability Probability and Playing Cards Probability for Rolling Two Dice Solved Probability Problems Probability for Rolling Three Dice 9th Grade Math From Probability for Rolling Three Dice to HOME PAGE
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Always treat the rolling of three dice as 3 separate rolls. Because in reality that is what happens: The dice will never land at the same time. If you took an action camera and slowed down the landing of the three dice you would see they would each finish at different times. Always treat the roll of three dice as three separate rolls. Because in reality that is what happens. Example If one throws three dice together on to the ground what is the probability of getting three 4s? Use `Probability\ =(The\ \n\umber\ of\ ways\ of \ ac\hiev\i\ng\ suc\ess)/(T\he\ \t\otal\ n\umber\ of \ possibl\e\ outcomes` And Always draw a probability tree. `Roll\i\ng\ a\ 4\ on\ a\ dice,\ probability=1/6\ \overset{(Right)}{\underset{(All\ possibilites)}{text}} ]` Probability three number 4s = `1/6\ times\1/6\ times1/6=1/(6\ times6\ times6)=1/(216)` Probability = 1 in 216 chance or = 0.004630 And if 1 = 100% then 0.004630=0.46% chance Answer: If one throws three dice together on the ground the probability of getting three 4s is 0.004630. More Info
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