Consider the game of independently throwing three fair six-sided dice. there are six combi- nations in which the three resulting numbers can sum up to 9, and also six combinations in which they can sum up to 10: 9 = 1 + 2 + 6 =1+3+5 = 1+4+4 = 2+ 3+ 4 = 2+2 +5 = 3+3+3 10 = 1+3+ 6 = 1+ 4+ 5 = 2+2+6 = 2+3+5 = 2 + 4 + 4 = 3+3+4 this seems to suggest that the chances of throwing 9 and 10 should be equal. yet, a certain gambler in florence in the early xvii century (most likely, the grand duke of tuscany cosimo ii de medici) noticed that in practice it is more likely to throw 10 than 9. show that the probability to get a total of 10 is, in fact, larger than the probability to get a total of 9. (this problem was solved for the duke by galileo galilei.)

See explanation below.

Step-by-step explanation:

1) First let's take a look at the combinations that sum up 10:

Notice that when we have 3 different numbers on the dice, we can permute them in 6 different ways. For example: Let's take 1 + 3 + 6, we can get this sum with these permutations:

1 + 3 + 6, 1 + 6 + 3, 3 + 6 + 1, 3 + 1 + 6, 6 + 1 + 3, 6 + 3 + 1.

And when we have two different numbers on the dice, we can permute them in 3 different ways:

2 + 2 + 6, 2 + 6 +2, 6 + 2 + 2.

So now we're going to write down the 6 combinations that sum up 10 but we're going to write down how many permutations of them we get:

Total of permutations: 6 + 6 + 3 + 6 + 3 + 3 =27.

Thus we have 27 different ways of getting a sum of 10.

2) Now we're going to take a look at the combinations that sum up 9 and we're going to proceed in a similar way:

Total of permutations: 6 + 6 + 3 + 6 +3 + 1 = 25.

Thus we have 25 different ways of getting a sum of 10

And we can conclude that the probability of getting a total of 10 is larger than the probability to get a total of 9.

i can't solve unless i know what the line is

step-by-step explanation:

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