Exercise 1: Conditional Probability and an Independence Check — Possible Solution
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GIVEN
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S = {1,2,3,4,5,6}
G = "rolling a number <= 3" = {1, 2, 3}
H = "rolling an even number" = {2, 4, 6}

STEP 1: THE OVERLAP
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G ∩ H = {2}  (the only outcome satisfying both conditions)
P(G ∩ H) = 1/6

STEP 2: P(G) AND P(H)
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P(G) = 3/6 = 0.5
P(H) = 3/6 = 0.5

STEP 3: P(G|H)
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P(G|H) = P(G ∩ H) / P(H) = (1/6) / (1/2) = 1/3

Given the roll is even (2, 4, or 6), only 1 of those 3 outcomes (the
2) also satisfies "<= 3" - matching 1/3 directly.

STEP 4: P(H|G)
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P(H|G) = P(G ∩ H) / P(G) = (1/6) / (1/2) = 1/3

Given the roll is <= 3 (1, 2, or 3), only 1 of those 3 outcomes (the
2) is also even - also matching 1/3 directly. (These two conditional
probabilities happen to be numerically equal here, but that's a
coincidence of this specific example, not a general rule - per this
chapter's own warning, P(A|B) and P(B|A) generally differ.)

STEP 5: THE INDEPENDENCE TEST
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P(G) x P(H) = 0.5 x 0.5 = 0.25
P(G ∩ H) = 1/6 ~= 0.1667

Since 0.25 does not equal 1/6, G and H are NOT independent - they are
dependent events. Knowing the roll is even genuinely changes the
probability it's also <= 3 (and vice versa).

WHY THIS WORKS AS AN ANSWER
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Both conditional probabilities are computed directly from this
chapter's own formula using the correct overlap and denominator for
each direction, and independence is determined by this chapter's own
multiplication-rule test rather than assumed from the fact that
P(G|H) happens to equal P(H|G) in this particular case.