04-BS-16 · May 2013

Question 8 of 12: Discrete Probability — Dice and Seating

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Examination, 04-BS-16 Discrete Mathematics, May 2013. Closed book, no aids, 3 hours, 12 questions of 10 marks each (100 marks); the exam instructs "answer 10 of 12" but every question is solved below as a complete study resource.

Reference texts: Rosen, Discrete Mathematics and Its Applications, 7th ed. (logic Ch.1, induction & recursion Ch.5, counting Ch.6, discrete probability Ch.7, relations Ch.9, graphs Ch.10-11); Epp, Discrete Mathematics with Applications.

Question 8: Discrete Probability — Dice and Seating (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Three fair 6-sided dice (216 equally-likely outcomes). Three twin pairs (6 people) randomly assigned to 6 seats arranged as 3 adjacent pairs.

Find. (a) $P(\text{all same})$, $P(\text{all different})$, $P(\text{at least one six})$. (b) $P(\text{every twin pair sits together}).$

(a-i) All three dice the same value. Sample space $6^3=216$. Favourable: value chosen 6 ways, all three dice match it: $6$ outcomes. $$P = \frac{6}{216} = \boxed{\frac{1}{36}}.$$
(a-ii) All three dice different values. Choose the first die's value (6 ways), second must differ (5 ways), third must differ from both (4 ways): $6\times5\times4=120$. $$P = \frac{120}{216} = \boxed{\frac{5}{9}}.$$
(a-iii) At least one six. Complement: no die shows six, each die has 5 non-six faces: $5^3=125$. $$P(\text{at least one six}) = 1-\frac{125}{216} = \boxed{\frac{91}{216}}.$$
(b) Twin-seating probability. Total assignments of 6 distinct people to 6 seats: $6!=720$. For everyone to sit beside their own twin: assign the 3 twin PAIRS to the 3 seat-PAIRS ($3!=6$ ways), then within each seat-pair the two twins can sit left/right in $2$ ways each, giving $2^3=8$. Favourable count $=3!\times2^3=6\times8=48$. $$P = \frac{48}{720} = \boxed{\frac{1}{15}}.$$

Final results — Question 8
Part	Result
(a-i) All same	$1/36$
(a-ii) All different	$5/9$
(a-iii) At least one six	$91/216$
(b) Twin seating	$1/15$

Topic: Classical (equally-likely-outcomes) probability; the complement rule; multiplicative counting for structured random assignments.

Key relations: $P(E)=\dfrac{|E|}{|S|}$ for equally-likely outcomes; $P(\text{at least one})=1-P(\text{none})$; independent multiplicative choices multiply counts (pair-to-seat-pair assignment $\times$ within-pair ordering).

Why this works: Each die roll is independent and uniform over 6 faces, so counting favourable ordered triples directly (or via their complement) is exact. The twin-seating problem decomposes into two independent choices — WHICH twin pair occupies WHICH seat-pair, and WHICH twin of each pair sits on the left vs. right — and the counting principle multiplies these independent choice-counts together.

Common pitfall: Using unordered outcomes (36 combinations of two dice, or similar) instead of the full $216$ ordered sample space, which breaks the equally-likely assumption for three dice; forgetting the factor $2^3$ (which twin sits on which side) in the seating problem.

Source: Rosen, Discrete Mathematics and Its Applications, §7.1-7.2 (probability theory, probability of complements and unions).

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