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The probability for each value is $\frac{1}{9}$.

The probability of each event is given as $\frac{1}{9}$.

$$\begin{gathered} P\left(Y=7\right)=\frac{1}{9} \\ P\left(Y=8\right)=\frac{1}{9} \\ P\left(Y=9\right)=\frac{1}{9} \end{gathered}$$

It can not be possible for the 1^st digit to be 2 different kinds of digits at the same time, therefore the above events are mutually exclusive.

Apply the addition rule for events of a mutually exclusive nature:

$$\begin{gathered} P\left(Y>6\right)=P\left(Y=7\right)+P\left(Y=8\right)+P\left(Y=9\right) \\ =\frac{1}{9}+\frac{1}{9}+\frac{1}{9} \\ =\frac{3}{9} \\ =\frac{1}{3}≈33.33\% \end{gathered}$$

Benford's Law:

$$\begin{gathered} P\left(Y=7\right)=0.058 \\ P\left(Y=8\right)=0.051 \\ P\left(Y=9\right)=0.046 \end{gathered}$$

It can not be possible for the 1^st digit to be 2 different kinds of digits at the same time, therefore the above events are mutually exclusive.

Apply the addition rule for events of a mutually exclusive nature:

$$\begin{gathered} P\left(Y>6\right)=P\left(Y=7\right)+P\left(Y=8\right)+P\left(Y=9\right) \\ =0.058+0.051+0.046 \\ =0.155 \\ =15.5\% \end{gathered}$$

If the fraction of the first digit greater than 6 is closer to 0.3333 than 0.155, the cost report is likely to be fraudulent.

$$\begin{gathered} \mathrm{μ}=∑xP\left(X=x\right) \\ =1×\frac{1}{9}+2×\frac{1}{9}+3×\frac{1}{9}+4×\frac{1}{9}+5×\frac{1}{9}+6×\frac{1}{9}+7×\frac{1}{9}+8×\frac{1}{9}+9×\frac{1}{9} \\ =\frac{1+2+3+4+5+6+7+8+9}{9} \\ =\frac{45}{9} \\ =5 \end{gathered}$$

In the following graphic, we can see that the anticipated value is 5 because the distribution is symmetric. The mean sits exactly in the middle of the distribution for a symmetric distribution, implying that the mean must be 5.

The uniform distribution's mean is 5, and Benford's law's mean is 3.441.

If the average of the first digits in an expense report is less than 3.441, the expense report is likely to be fake.

As a result, if we get an expense report with a mean of all initial digits that is closer to 5 than 3.441, we'll conclude it's a fake.