Question

faked numbers on tax returns, invoices, or expense account claims often display patterns that aren’t present in legitimate records. some patterns, like too many round numbers, are obvious and easily avoided by a clever crook. others are more subtle, like a tendency that the first digits of numbers in legitimate records often follow a model known as benford’s law. call the first digit of a randomly chosen record x for short. benford’s law gives this probability model for x (note that a first digit can’t be 0).

first digit: 1 2 3 4 5 6 7 8 9
probability: 0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046

(a) write the event “first digit is a 1” in terms of x. then find its probability.
options:

• p(x = 1) = 0.699
• p(x > 1) = 0.699
• p(x > 1) = 0.301
• p(x = 1) = 0.301

(b) what’s the probability the first digit of a randomly chosen record is at least 6?
(include 3 decimal places.)

Explanation:

Part (a)

Step1: Define the event

The event "first digit is a 1" is \( P(X = 1) \). From the table, the probability for \( X = 1 \) is given as \( 0.301 \)? Wait, no, wait the table: Wait, the first digit 1 has probability 0.301? Wait, no, the table shows first digit 1: 0.301? Wait, no, looking at the table, first digit 1: 0.301? Wait, no, the options: the first option is \( P(X = 1)=0.699 \) (no, wait the options are: \( P(X = 1)=0.699 \), \( P(X = 1)=0.301 \), \( P(X > 1)=0.699 \), \( P(X > 1)=0.301 \). Wait, the total probability should sum to 1. The probability for \( X = 1 \) is 0.301? Wait, no, the table: first digit 1: 0.301? Wait, no, the table's first digit 1 has probability 0.301? Wait, no, let's check the total. Wait, the probabilities for digits 1 - 9: 0.301 (1), 0.176 (2), 0.125 (3), 0.097 (4), 0.079 (5), 0.067 (6), 0.058 (7), 0.051 (8), 0.046 (9). Wait, no, the first digit 1's probability is 0.301? Wait, no, the sum: 0.301 + 0.176 = 0.477, +0.125 = 0.602, +0.097 = 0.699, +0.079 = 0.778, +0.067 = 0.845, +0.058 = 0.903, +0.051 = 0.954, +0.046 = 1.0. Ah, so \( P(X = 1) = 0.301 \), and \( P(X > 1) = 1 - P(X = 1) = 1 - 0.301 = 0.699 \). Wait, but the event "first digit is a 1" is \( X = 1 \), so \( P(X = 1) = 0.301 \)? Wait, no, the options: the first option is \( P(X = 1)=0.699 \) (wrong), second is \( P(X = 1)=0.301 \) (correct), third is \( P(X > 1)=0.699 \) (correct for \( X > 1 \)), fourth is \( P(X > 1)=0.301 \) (wrong). Wait, the question is "Write the event 'first digit is a 1' in terms of X. Then find its probability." So the event is \( X = 1 \), and its probability is 0.301? Wait, no, the sum of probabilities: 0.301 (1) + 0.176 (2) + 0.125 (3) + 0.097 (4) + 0.079 (5) + 0.067 (6) + 0.058 (7) + 0.051 (8) + 0.046 (9) = 0.301+0.176=0.477+0.125=0.602+0.097=0.699+0.079=0.778+0.067=0.845+0.058=0.903+0.051=0.954+0.046=1.0. So \( P(X = 1) = 0.301 \), so the correct option is \( P(X = 1) = 0.301 \).

Step2: Confirm the probability

The event "first digit is 1" is \( X = 1 \). From the probability distribution table, \( P(X = 1) = 0.301 \).

Step1: Identify the event

The event "first digit is at least 6" means \( X \geq 6 \), which includes \( X = 6, 7, 8, 9 \).

Step2: Sum the probabilities

From the table, the probabilities for \( X = 6, 7, 8, 9 \) are \( 0.067 \), \( 0.058 \), \( 0.051 \), and \( 0.046 \) respectively.

Step3: Calculate the sum

\( P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) \)
\( = 0.067 + 0.058 + 0.051 + 0.046 \)
First, \( 0.067 + 0.058 = 0.125 \)
Then, \( 0.125 + 0.051 = 0.176 \)
Then, \( 0.176 + 0.046 = 0.222 \)

Wait, wait, the table's first digit 6: probability is 0.067? Wait, the original table (from the image) shows first digit 6: 0.067? Wait, the user's table: first digit 6: 0.067, 7: 0.058, 8: 0.051, 9: 0.046. So sum: 0.067 + 0.058 = 0.125; 0.125 + 0.051 = 0.176; 0.176 + 0.046 = 0.222. Rounded to 3 decimal places, 0.222.

Answer:

(for part a):
\( P(X = 1) = 0.301 \) (corresponding option: the one with \( P(X = 1) = 0.301 \))

Part (b)