Exercise 3: Odd Numbers, Numbers Less Than 4, and the Double-Counting Mistake — Possible Solution
====================================================================================

GIVEN
------------------------------
Rolling a single die, S = {1,2,3,4,5,6}
E = "rolling an odd number" = {1, 3, 5}
F = "rolling a number less than 4" = {1, 2, 3}

STEP 1: P(E) AND P(F)
------------------------------
P(E) = 3/6 = 0.5
P(F) = 3/6 = 0.5

STEP 2: P(E AND F)
------------------------------
E and F = {1, 3, 5} and {1, 2, 3} share the outcomes 1 and 3.

E ∩ F = {1, 3}
P(E and F) = 2/6 = 1/3

STEP 3: THE UNION RULE
------------------------------
P(E or F) = P(E) + P(F) - P(E and F)
          = 0.5 + 0.5 - 1/3
          = 1 - 1/3
          = 2/3

Direct check: E ∪ F = {1, 2, 3, 5}, which has 4 outcomes out of 6:
4/6 = 2/3, matching exactly.

STEP 4: THE NAIVE (INCORRECT) SUM
------------------------------
P(E) + P(F) = 0.5 + 0.5 = 1.0

This naive sum claims a 100% (certain) probability of rolling either
an odd number or a number less than 4 - but rolling a 4 or a 6
satisfies neither condition, so certainty is obviously wrong.

WHICH OUTCOMES GET DOUBLE-COUNTED
------------------------------
The naive sum counts every outcome in E once and every outcome in F
once, separately. The outcomes 1 and 3 belong to BOTH E and F, so
they each get counted twice in the naive sum (once as members of E,
once again as members of F) instead of once each, as they should be
in a true probability. That double-counted excess is exactly
P(E and F) = 1/3, which is precisely the amount the union rule
subtracts back out: 1.0 - 1/3 = 2/3, the correct answer.

WHY THIS WORKS AS AN ANSWER
------------------------------
The union rule is applied correctly using this chapter's own formula
and cross-checked against a direct count of E ∪ F, and the naive sum's
error is explained by identifying exactly which outcomes (1 and 3)
get counted twice and showing that the excess exactly equals the
P(E and F) term the union rule subtracts.