Benford’s Law – Using a probability model and identifying outcomes
A = {first digit is 1}
B = {first digit is 6 or greater}
C = {first digit is odd}
D = {first digit is even}
E = {first digit is at most 4}
F = {first digit is at least 2}
P(A) = P(1) = 0.301
P(B) = P({6,7,8,9}) = P(6) + P(7) + P(8) + P(9) = 0.222
P(BC) = P({1,2,3,4,5}) = P(1) + P(2) + P(3) + P(4) + P(5) = 0.778
Note … the notation BC denotes the “complement of B” and refers to all outcomes that are not included
in event B. Since the union of B and BC must be the entire sample space, one way to calculate P(BC) is to
use the relationship that P(BC) + P(B) = 1, in other words, use P(BC) = 1 – P(B).
= P(A and B) = P({ }) = 0 (This is the “empty set,” i.e., a digit can’t be both 1 and 6 or more!)
= P(A or B) = P({1,6,7,8,9}) = P(A) + P(B) = 0.523 (there is no overlap between A and B)
P(C) = P({1,3,5,7,9}) = P(1) + P(3) + P(5) + P(7) + P(9) = 0.609
= P(A and C) = P(1) = 0.301 (Only 1 is both 1 and odd!)
= P(A or C) = P({1,3,5,7,9}) = P(1) + P(3) + P(5) + P(7) + P(9) = 0.609
= P(B and D) = P({6,8}) = P(6) + P(8) = 0.118
= P(B or D) = P({2,4,6,7,8,9}) = P(2) + P(4) + P(6) + P(7) + P(8) + P(9) = 0.495
P(E) = P({1,2,3,4}) = P(1) + P(2) +P(3) + P(4) = 0.699
P(F) = P({2,3,4,5,6,7,8,9}) = P(2) + P(3) + … + P(9) = 0.699
But wait … doesn’t F consist of every outcome except for “The first digit is 1”? Then we can just
compute P(1) and subtract it from the total probability of 1. In symbols:
P(F) = 1 – P(1) = 1 - .301 = 0.699. So in this case, the events “the first digit is at least 2” and “the first
digit is 1” are complements …
Independence and the multiplication rule
Toss a fair coin twice …: S = {HH, HT, TT, TH}
A = first toss is a head
B = second toss is a head
P(A) = P({HH, HT}) = 0.5
P(B) = P({HT, TT}) = 0.5
P(A and B) = P({HH}) = 0.5 x 0.5 = 0.25 (The fact that the first toss is a head has no impact on whether or not the
second toss is a head, so we can just multiply their individual probabilities.) We say that the tosses are
independent of one another.
Now … draw two cards from the same deck of 52 cards, without replacement.
A = first card drawn is red
B = second card drawn is red
P(A) = 26/52
P(B) = 25/51 if the first card was red, or 26/51 if the first card was not red.
Either way, since the probability of B occurring depends on whether or not A occurred, events A and B are NOT
independent, and so we cannot simply multiply their probabilities.
Make sure you read and study example 6.14 on page 353. It makes use of independence and of the
complement!!