Hypergeometric and Binomial Distributions: 04-BS-2, December 2014, Q3

Question 3: Hypergeometric and Binomial Distributions (20 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) 16 laptops: 10 Amon, 6 King Tut; a sample of 6 drawn without replacement. (B) 80% of members are happy with the posted speed limits.

Find. (A)(a) P(at least 4 Amon in 6); (A)(b) pmf of Y = King Tut count; (B)(a) P(4<X<8 happy of 15); (B)(b) P(>380 unhappy of 2,000).

Approach. Sampling 6 laptops without replacement from a fixed lot of two types is Hypergeometric; each Truck Drivers Association member is an independent Bernoulli(0.80) trial, so a sample of 15 is Binomial exactly, while a sample of 2,000 is well approximated by the Normal with a continuity correction.

(A)(a) Hypergeometric tail probability. Let X = number of Amon laptops in the sample of 6, drawn from N=16 (K=10 Amon, N−K=6 King Tut): $$P(X=x)=\dfrac{\binom{10}{x}\binom{6}{6-x}}{\binom{16}{6}},\qquad x=0,\dots,6.$$ "At least four Amon" is $X\ge4$: $$P(X\ge4)=P(4)+P(5)+P(6)=0.3934+0.1888+0.0262=\boxed{0.6084}.$$
(A)(b) pmf of the King Tut count Y. Y=6−X, drawn from N=16 with K=6 King Tut laptops: $$P(Y=y)=\dfrac{\binom{6}{y}\binom{10}{6-y}}{\binom{16}{6}},\qquad y=0,1,\dots,6.$$

pmf of Y (King Tut count in the sample of 6)
y	0	1	2	3	4	5	6
P(Y=y)	0.0262	0.1888	0.3934	0.2997	0.0843	0.0075	0.0001

(B)(a) Binomial probability, strict interval. "More than four but fewer than eight" means $5\le X\le7$, so with $X\sim\text{Binomial}(15,0.80)$: $$P(5\le X\le7)=\sum_{k=5}^{7}\binom{15}{k}(0.8)^k(0.2)^{15-k}=\boxed{0.00423}.$$ This is small because with p=0.80 the Binomial(15,0.80) distribution is centred at 12, far above the 5–7 window; the mass in that low tail is a genuine small-probability event, not a modelling error.
(B)(b) Normal approximation to a large Binomial. Let Y = number UNHAPPY among 2,000 members, so $Y\sim\text{Binomial}(2000,\,0.20)$ (since 80% happy ⇒ 20% unhappy), with $\mu_Y=np=400$ and $\sigma_Y=\sqrt{np(1-p)}=\sqrt{2000(0.2)(0.8)}=17.889$. Applying the continuity correction, "more than 380" becomes $Y\ge381$: $$P(Y>380)\approx1-\Phi\!\left(\dfrac{380.5-400}{17.889}\right)=1-\Phi(-1.090)=\Phi(1.090)=\boxed{0.8622}.$$

Final results — Question 3
Part	Quantity	Result
(A)(a)	P(Amon ≥ 4 of 6)	0.6084
(A)(b)	pmf of Y (King Tut count)	see table above
(B)(a)	P(5 ≤ happy ≤ 7 of 15)	0.00423
(B)(b)	P(unhappy > 380 of 2,000)	0.8622

Topic: Hypergeometric sampling without replacement vs. Binomial sampling of independent Bernoulli trials, and the Normal approximation to a large Binomial.

Key relations: Hypergeometric $P(X=x)=\binom{K}{x}\binom{N-K}{n-x}/\binom{N}{n}$; Binomial $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$; Normal approx. $X\approx N(np,\,np(1-p))$ with continuity correction for large n.

Why this works: a small fixed lot sampled without replacement changes the composition remaining after each draw (Hypergeometric), whereas each Truck Drivers Association member is effectively an independent trial drawn from a much larger population, so their happy/unhappy status is well modelled as i.i.d. Bernoulli (Binomial).

Common pitfall: using the Binomial pmf for the laptop draw (ignoring that the population shrinks as laptops are removed), and forgetting the continuity correction (or the happy/unhappy complement) when Normal-approximating a large Binomial tail.

Source: Walpole, Myers, Myers & Ye, Ch. 5 (Discrete distributions and their approximations).

Question 3 of 8: Hypergeometric and Binomial Distributions

Question 3: Hypergeometric and Binomial Distributions (20 marks)