04-BS-16 · May 2013

Question 12 of 12: Conditional Probability and the Binomial Distribution

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 12: Conditional Probability and the Binomial Distribution (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. (a) 3-child family, 8 equally-likely B/G sequences. (b) 8 independent fair bits (each 0 or 1 with probability $\tfrac12$).

Find. (a) $P(\text{exactly 2 boys, 1 girl}\mid\text{at least 1 boy})$. (b)(i) $P(\text{exactly 3 ones})$. (b)(ii) $P(\text{at least 3 ones})$.

(a) Reduce the sample space. "At least one boy" excludes only the sequence GGG, leaving $8-1=7$ equally-likely sequences.
(a) Count the favourable sequences. Sequences with exactly 2 boys and 1 girl: BBG, BGB, GBB — 3 sequences (all lie within the reduced 7-sequence space, since each contains at least one boy). $$P(\text{2B,1G}\mid\ge1\text{ boy}) = \frac{3}{7} = \boxed{\frac{3}{7}}.$$
(b-i) Exactly three 1's among 8 bits. Choose which 3 of the 8 bit positions are 1 ($\binom{8}{3}$ ways); each specific pattern has probability $(\tfrac12)^8$: $$P(\text{exactly 3 ones}) = \binom{8}{3}\left(\frac12\right)^8 = \frac{56}{256} = \boxed{\frac{7}{32}}.$$
(b-ii) At least three 1's. Complement: fewer than 3 ones, i.e. 0, 1, or 2 ones. $$P(0)+P(1)+P(2) = \frac{\binom{8}{0}+\binom{8}{1}+\binom{8}{2}}{2^8} = \frac{1+8+28}{256} = \frac{37}{256}.$$ $$P(\text{at least 3 ones}) = 1 - \frac{37}{256} = \boxed{\frac{219}{256}}.$$

Final results — Question 12
Part	Result
(a) $P(\text{2B,1G}\mid\ge1\text{B})$	$3/7$
(b-i) $P(\text{exactly 3 ones})$	$7/32$
(b-ii) $P(\text{at least 3 ones})$	$219/256$

Topic: Conditional probability with a reduced sample space; the binomial probability distribution.

Key relations: Conditional probability on equally-likely outcomes: $P(E\mid F)=\dfrac{|E\cap F|}{|F|}$ (restrict to $F$, then count $E$ within it); Binomial: $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$, here $n=8,p=\tfrac12$; complement rule for "at least."

Why this works: Because all 8 original sequences are equally likely, conditioning on "at least one boy" simply restricts attention to the 7 sequences that satisfy it and renormalizes; every one of the 3 favourable "2 boys 1 girl" sequences happens to already contain a boy, so none is lost in the restriction. Each bit is an independent fair coin, so the byte is a classic Binomial$(8,\tfrac12)$ random variable, and summing $P(0),P(1),P(2)$ is cheaper than summing $P(3)$ through $P(8)$ directly.

Common pitfall: Computing $P(\text{2B,1G})$ over the ORIGINAL 8-sequence space (giving $3/8$) instead of correctly conditioning on the reduced 7-sequence space; miscounting $\binom{8}{2}=28$ (a common slip is $\binom{8}{2}=16$).

Source: Rosen, Discrete Mathematics and Its Applications, §7.2 (conditional probability), §7.4 (the binomial distribution / Bernoulli trials).

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