Roland2rule wrote:If four dice are thrown together, then the probability that the sum on them together is either 19 or 23
1. 4/108
2. 12/108
3. 9/108
4. 5/108
5. 6/108
Notice that the probability of getting a sequence of 4 numbers in a specific order, e.g., 1-2-3-4, is 1/6 x 1/6 x 1/6 x 1/6 = 1/6^4.
Now let's list all the groups of 4 numbers that add up to 19 with the number of ways the numbers within a group can be arranged in parenthesis:
6-6-6-1 (4!/3! = 4)
6-6-5-2 (4!/2! = 12)
6-6-4-3 (4!/2! = 12)
6-5-5-3 (4!/2! = 12)
6-5-4-4 (4!/2! = 12)
5-5-5-4 (4!/3! = 4)
Thus the probability of getting a sum of 19 is (4 x 2 + 12 x 4) x 1/6^4 = 56/6^4.
Similarly, let's list all the groups of 4 numbers that add up to 23 with the number of ways the numbers within a group can be arranged in parenthesis:
6-6-6-5 (4!/3! = 4)
We see that there is only one group of 4 numbers that add up to 23 and within this group, there are 4 ways to arrange them. Thus the probability of getting a sum of 23 is 4 x 1/6^4 = 4/6^4.
Thus the probability of getting a sum of 19 or 23 is 56/6^4 + 4/6^4 = 60/6^4 = (6 x 10)/6^4 = 10/6^3 = 10/216 = 5/108.
Answer:
D
