Probability a die shows even or greater than three

Question

You roll a single fair six-sided die. What is the probability the result is even or greater than three?

Accepted Answer

The even outcomes are \{2, 4, 6\}, three faces. The outcomes greater than three are \{4, 5, 6\}, three faces. Adding 	frac{3}{6} + 	frac{3}{6} double-counts the faces in both sets, namely \{4, 6\}, which are even and above three. Inclusion-exclusion: P(E \cup G) = P(E) + P(G) - P(E \cap G) = \frac{3}{6} + \frac{3}{6} - \frac{2}{6} = \frac{4}{6} = \frac{2}{3} \approx 0.667. Listing the favorable faces directly confirms it: \{2, 4, 5, 6\} is four out of six. Takeaway. P(A \cup B) = P(A) + P(B) - P(A \cap B). Adding is only valid for disjoint events; when the two conditions share outcomes (her…

Probability a die shows even or greater than three

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