Question

7. Benford’s law Refer to Exercise 5. The first digit of a randomly chosen expense account claim follows Benford’s law. Consider the events A = first digit is 7 or greater and B = first digit is odd.

(a) What outcomes make up the event A? What is P(A)?

(b) What outcomes make up the event B? What is P(B)?

(c) What outcomes make up the event “A or B”? What is P(A or B)? Why is this probability not equal to P(A) + P(B)?

Short answer

a)The outcomes that make up the event A is $0.155$

b)The probability of $P\mathit{\left(}B\mathit{\right)}$is $0.609$

c)$P\mathit{\left(}A\mathit{\text{or}}B\mathit{\right)}\mathit{\ne }P\mathit{\left(}A\mathit{\right)}\mathit{+}P\mathit{\left(}B\mathit{\right)}\text{because the events are not mutually exclusive.}$

Step-by-step solution

Part (a) Step 1: Given Information

Given that the probability distribution $\begin{array}{|lrrrrrrrrr|}\hline X& 1& 2& 3& 4& 5& 6& 7& 8& 9\\ \text{Probability}& 0.301& 0.176& 0.125& 0.097& 0.079& 0.067& 0.058& 0.051& 0.046\\ \hline\end{array}$

Part (a) Step 2:  Calculation

According to the information, consider $A$as an event indicating that the first digit is $7$or higher The following are the consequences of event A:

$\text{Outcomes}=\{7,8,9\}$

$P\left(A\right)$can be calculated as:

localid="1649993368336" $$\begin{gathered} P\left(A\right)=P(X=7)+P(X=8)+P(X=9) \\ =0.058+0.051+0.046 \\ =0.155 \end{gathered}$$

Part (b) Step 1: Given Information

Given that the probability distribution

$\begin{array}{|lcccccccc|}\hline \text{Days}& 0& 1& 2& 3& 4& 5& 6& 7\\ \text{Probability}& 0.68& 0.05& 0.07& 0.08& 0.05& 0.04& 0.01& 0.02\\ \hline\end{array}$

Part (b) Step 2: Calculation

If $B$is the event indicating that the first digit is odd, The results of event$B$can be written as follows:

$\text{Outcomes}=\{1,3,5,7,9\}$

$\mathit{P}\left(\mathit{B}\right)$can be calculated as:

localid="1649993375643" $$\begin{gathered} P\left(B\right)=P(X=1)+P(X=3)+P(X=5)+P(X=7)+P(X=9) \\ =0.301+0.125+0.079+0.058+0.046 \\ =0.609 \end{gathered}$$

Part (c) Step 1: Given Information

Given that the probability distribution

$\begin{array}{|lcccccccc|}\hline \text{Days}& 0& 1& 2& 3& 4& 5& 6& 7\\ \text{Probability}& 0.68& 0.05& 0.07& 0.08& 0.05& 0.04& 0.01& 0.02\\ \hline\end{array}$

Part (c) Step 2: Calculation 

The outcomes for events $A$and $B$can be written as follows, using the data from the previous sections:

$\text{Outcomes}=\{1,3,5,7,8,9\}$

$\mathit{P}(\mathit{A}$or $B)$can be calculated as:

localid="1649993383176" $$\begin{gathered} P(AorB)=P\mathit{(}X\mathit{=}\mathit{1}\mathit{)}\mathit{+}P\mathit{(}X\mathit{=}\mathit{3}\mathit{)}\mathit{+}P\mathit{(}X\mathit{=}\mathit{5}\mathit{)}\mathit{+}P\mathit{(}X\mathit{=}\mathit{7}\mathit{)}\mathit{+}P\mathit{(}X\mathit{=}\mathit{8}\mathit{)}\mathit{+}P\mathit{(}X\mathit{=}\mathit{9}\mathit{)} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\mathit{0.301}\mathit{+}\mathit{0.125}\mathit{+}\mathit{0.079}\mathit{+}\mathit{0.058}\mathit{+}\mathit{0.051}\mathit{+}\mathit{0.046} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\mathit{0.660} \end{gathered}$$

localid="1649993386761" $\text{Here,}P\left(A\text{or}B\right)\ne P\left(A\right)+P\left(B\right)\text{because the events are not mutually exclusive.}$