Worked solution
Count five events for two ordinary dice
Two ordinary dice are thrown. Find the probability that (i) their sum is 3, (ii) their sum exceeds 9, (iii) they show the same number, (iv) their numbers differ by more than 2, (v) their product is even.
CJC · Lecture · Self-Practice 4
(i) Sum three
Among 36 equally likely ordered pairs, the favourable outcomes are (1,2),(2,1). Probability 2/36=1/18.
(ii) Sum greater than nine
Sums 10,11,12 contribute 3,2,1 ordered pairs respectively. The six outcomes are (4,6),(5,5),(6,4),(5,6),(6,5),(6,6). Probability 6/36=1/6.
(iii) Matching numbers
The six diagonal outcomes(1,1)through (6,6)give 6/36=1/6.
(iv) Absolute difference greater than two
Differences 3,4,5 give 6,4,2 ordered outcomes respectively, including both orientations. Thus(6+4+2)/36=12/36=1/3.
(v) Even product
The product is odd only when both dice are odd. There are 3 odd choices for each die, so 9 odd-product outcomes. The complement gives 1−9/36=27/36=3/4. All five printed answers 1/18,1/6,1/6,1/3,3/4 are retained; the derivations supplement the answer-only source.
Common mistakes
Treat the two dice as distinguishable ordered outcomes and assume independent fair dice. “Differ by more than 2” means absolute difference, so include both orders.