Dear Ravi, The solution for this is as follows:- As, three dice are thrown simultaneously. So , The combinations that result in getting sum of the digits of 15 are: (5, 5, 5), (3, 6, 6), (4, 5, 6) Total number of permutations for the above combinations: 1+3!/2! +3! =10 Desirable outcomes=10 Total number of outcomes if 3 dice are thrown: 6 x 6 × 6 = 216. Therefore, Probability= 10/216= 5/108. Thus, Answer is 5/108 Hope this sorted your query. All the Best!!
Dice provide great illustrations for concepts in probability. The most commonly used dice are cubes with six sides. Here, we will see how to calculate probabilities for rolling three standard dice. It is a relatively standard problem to calculate the probability of the sum obtained by rolling two dice. There are a total of 36 different rolls with two dice, with any sum from 2 to 12 possible. How does the problem change if we add more dice? Just as one die has six outcomes and two dice have 62 = 36 outcomes, the probability experiment of rolling three dice has 63 = 216 outcomes. This idea generalizes further for more dice. If we roll n dice then there are 6n outcomes. We can also consider the possible sums from rolling several dice. The smallest possible sum occurs when all of the dice are the smallest, or one each. This gives a sum of three when we are rolling three dice. The greatest number on a die is six, which means that the greatest possible sum occurs when all three dice are sixes. The sum of this situation is 18. When n dice are rolled, the least possible sum is n and the greatest possible sum is 6n.
As discussed above, for three dice the possible sums include every number from three to 18. The probabilities can be calculated by using counting strategies and recognizing that we are looking for ways to partition a number into exactly three whole numbers. For example, the only way to obtain a sum of three is 3 = 1 + 1 + 1. Since each die is independent from the others, a sum such as four can be obtained in three different ways:
Further counting arguments can be used to find the number of ways of forming the other sums. The partitions for each sum follow:
When three different numbers form the partition, such as 7 = 1 + 2 + 4, there are 3! (3x2x1) different ways of permuting these numbers. So this would count toward three outcomes in the sample space. When two different numbers form the partition, then there are three different ways of permuting these numbers. We divide the total number of ways to obtain each sum by the total number of outcomes in the sample space, or 216. The results are:
As can be seen, the extreme values of 3 and 18 are least probable. The sums that are exactly in the middle are the most probable. This corresponds to what was observed when two dice were rolled.
Text Solution Solution : Three dice are thrown <br> `n(S)=6^3=216` <br> A be the event of getting sum as `15` <br> `A = {(3, 6, 6), (4,6,5), (5,6,4), (6,6,3), (6,3,6), (6,4,5), (6,5,4), (4,5,6), (5,5,5), (5,4,6) }` <br> `P(A)=(10)/(216)` <br> If three dices are thrown simultaneously, then the number of all the possible outcomes are 63 = 216. i.e. number of favourable outcomes = n(A) = 10 Hence, required probability = P (getting a sum of 15) = \[\frac{10}{216}\] |