Dawson 201-SN1-RE tutoring

1. Count equally likely groups. The notation $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ counts the ways, without regard to order, to choose $k$ objects from $n$ distinct objects; the factorial $n!$ means the product of the positive integers through $n$, with $0!=1$. There are $\binom{12}{3}=12\cdot11\cdot10/(3\cdot2\cdot1)=220$ possible groups. 2. Count the desired groups. Exactly one defective component means choosing one from the $4$ defective components and two from the remaining $8$: $\binom41\binom82=4(28)=112$ groups. Writing $\Pr$ for probability, $\Pr(X=1)=112/220=28/55\approx0.5091$, or about $50.9\%$. This is sampling without replacement, often called the hypergeometric model. 3. Compute the expected count. Let $I_j$ be $1$ if draw $j$ is defective and $0$ otherwise, for $j=1,2,3$. Then $X=I_1+I_2+I_3$. Before seeing any outcomes, every position in the random draw order has defect probability $4/12$. Writing $\mathbb{E}$ for expected value, the rule for adding expected values, called linearity, gives $\mathbb{E}[X]=\mathbb{E}[I_1]+\mathbb{E}[I_2]+\mathbb{E}[I_3]=3(4/12)=1$. This addition rule does not require independent draws. An expected value of one is a long-run average, not a guarantee for any one sample. 4. Change the sampling model. With replacement and independent draws, the defect probability stays $p=4/12=1/3$. There are $\binom31=3$ positions for the single defective draw, and each such pattern has probability $p(1-p)^2$. Thus the new probability is $\binom31(1/3)(2/3)^2=4/9\approx0.4444$, or about $44.4\%$. This binomial formula would be invalid for the original dependent draws. Both models have expected count $1$, but different probabilities for individual counts.