Question 3 of 8: Binomial and Hypergeometric Sampling

Question 3: Binomial and Hypergeometric Sampling (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) $p=0.60$ pass rate, independent applicants (Binomial). (B) Lot of $N=16$ tanks with $K=4$ substandard, sample $n=5$ without replacement (Hypergeometric).

Find. (A)(a) $P(4\lt X\lt8)$, $n=14$; (A)(b) $P(X>8)$, $n=12$. (B)(a) $P(X\le2)$; (B)(b) full pmf of $X$ and $E(X)$.

Approach. Recognize independent-trial pass/fail as Binomial and finite-population without-replacement selection as Hypergeometric, then sum the pmf over the required outcomes.

1. (A)(a) Binomial, $n=14,\ p=0.6$. "More than 4 but fewer than 8" means $X\in\{5,6,7\}$. $$P(4\lt X\lt8)=\binom{14}{5}0.6^{5}0.4^{9}+\binom{14}{6}0.6^{6}0.4^{8}+\binom{14}{7}0.6^{7}0.4^{7}=0.0754+0.1281+0.1465=\boxed{0.2900}$$
2. (A)(b) Binomial, $n=12,\ p=0.6$. $P(X>8)=P(X=9)+P(X=10)+P(X=11)+P(X=12)$ $$=\binom{12}{9}0.6^9 0.4^3+\binom{12}{10}0.6^{10}0.4^2+\binom{12}{11}0.6^{11}0.4^1+\binom{12}{12}0.6^{12}0.4^0=\boxed{0.2253}$$
3. (B)(a) Hypergeometric, $N=16,\ K=4,\ n=5$. $$P(X\le2)=\sum_{k=0}^{2}\dfrac{\binom{4}{k}\binom{12}{5-k}}{\binom{16}{5}}=\dfrac{\binom{12}{5}}{\binom{16}{5}}+\dfrac{\binom{4}{1}\binom{12}{4}}{\binom{16}{5}}+\dfrac{\binom{4}{2}\binom{12}{3}}{\binom{16}{5}}=0.1813+0.4533+0.3022=\boxed{0.9368}$$
4. (B)(b) Full distribution and mean. $X$ ranges over $k=0,\dots,4$ (can't exceed the 4 substandard tanks or the sample size 5):

5. Mean. For a Hypergeometric r.v., $E(X)=n\dfrac{K}{N}=5\cdot\dfrac{4}{16}=\boxed{1.25}$ (matches $\sum k\,P(X=k)$ from the table above).