When a question like this is really fuzzy and you do not know from where.....think smart.

Now, we do know that in the first blank the question is

The probability that

at least two people will roll the same number.

From this problem is well-suited to the "1 - x” shortcut. You can calculate the probability of each of the four rolls resulting in a different number each time. In this scenario is way much easier to calculate the probability that you DO NO want and NOT those you wanna. Do not attack the problem up-front, but using the other way around.

The first roll is assigned a probability of l, since the first number that comes up will not be the same as that of any previous roll (because there has been no previous roll). In the next roll, the first number that came up must be excluded, so that there are 5 allowable outcomes. Likewise, the third roll will have 4 allowable outcomes, and the fourth roll will have 3 allowable outcomes. The overall probability of all four numbers being distinct is therefore equal to

\(1 * \frac{5}{6} * \frac{4}{6} * \frac{3}{6}= 1 - \frac{5}{18} = \frac{13}{18}\) =

0.72A is the answer

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