04-BS-16 · December 2013

Question 7 of 12: Discrete Probability — Dice and Conditional Probability

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

Notes on this paper

National Examination, 04-BS-16 Discrete Mathematics, Dec 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, sets Ch.2, induction & pigeonhole Ch.5-6, relations Ch.9, counting Ch.6, discrete probability Ch.7, graphs Ch.10-11); Epp, Discrete Mathematics with Applications.

Question 7: Discrete Probability — Dice and Conditional Probability (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. Two fair six-sided dice, 36 equally likely outcomes. Four-child family, 16 equally likely BGGB-type sequences.

Find. (a) Three dice probabilities (one unconditional pair, one conditional). (b) Two conditional probabilities of "four girls."

(a-i) Exactly one die shows six. Either die 1 is six and die 2 isn't, or vice versa: $2\times\frac16\times\frac56=\frac{10}{36}$: $$P = \boxed{\frac{5}{18}}.$$
(a-ii) Sum greater than three. Complement: sum $\le3$ occurs for $(1,1),(1,2),(2,1)$ — 3 outcomes out of 36, so $P(\text{sum}\le3)=\frac{3}{36}=\frac1{12}$: $$P(\text{sum}>3) = 1-\frac1{12} = \boxed{\frac{11}{12}}.$$
(a-iii) $P(\text{total}<7\mid\text{at least one die shows 2})$. "At least one 2" has $2\times6-1=11$ outcomes (rows/columns with a 2, minus the double-counted $(2,2)$). Among these, sum $<7$: $(2,1),(2,2),(2,3),(2,4),(1,2),(3,2),(4,2)$ — 7 outcomes: $$P = \boxed{\frac{7}{11}}.$$
(b-a) $P(\text{4 girls}\mid\text{youngest is a girl})$. "Youngest is a girl" fixes 1 of 4 positions, leaving $2^3=8$ equally likely sequences for the rest; only 1 of those 8 is all-girls (GGGG): $$P = \boxed{\frac18}.$$
(b-b) $P(\text{4 girls}\mid\text{at least one girl})$. "At least one girl" excludes only the all-boys sequence: $16-1=15$ sequences. Exactly 1 of those (GGGG) is all-girls: $$P = \boxed{\frac1{15}}.$$

Final results — Question 7
Part	Result
(a-i) Exactly one six	5/18
(a-ii) Sum > 3	11/12
(a-iii) Total < 7 \| at least one 2	7/11
(b-a) 4 girls \| youngest girl	1/8
(b-b) 4 girls \| at least 1 girl	1/15

Topic: Classical (equally-likely-outcomes) probability and conditional probability by direct outcome counting.

Key relations: $P(A\mid B)=\dfrac{|A\cap B|}{|B|}$ under a uniform sample space — no Bayes' rule needed when both numerator and denominator can be counted directly.

Why this works: Conditioning on "at least one die shows 2" (or "youngest is a girl") shrinks the sample space to only the outcomes consistent with the condition; the conditional probability is then just the fraction of THAT smaller space satisfying the target event — no need for the full Bayes formula when outcomes are equiprobable and both sets are easy to enumerate.

Common pitfall: Double-counting $(2,2)$ when computing "at least one die shows 2" (11, not 12, outcomes); confusing "youngest is a girl" (fixes one specific position) with "exactly one girl."

Source: Rosen, Discrete Mathematics and Its Applications, §7.1-7.2 (discrete probability, conditional probability).

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