Question

From a lot of 12 items containing 3 defectives, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If the variance of $X$ is $\frac{m}{n}$, where $\gcd(m, n) = 1$, then $n - m$ is equal to ________.

Answer 1

$a = 1 - \frac{\binom{3}{5}}{\binom{12}{5}}$
$b = 3 \cdot \frac{\binom{9}{4}}{\binom{12}{5}}$
$c = 3 \cdot \frac{\binom{9}{3}}{\binom{12}{5}}$
$d = 1 \cdot \frac{\binom{9}{2}}{\binom{12}{5}}$
$u = 0 \cdot a + 1 \cdot b + 2 \cdot c + 3 \cdot d = 1.25$
$\sigma^2 = 0 \cdot a + 1 \cdot b + 4 \cdot c + 9 \cdot d - u^2$
$\sigma^2 = \frac{105}{176}$
$\text{Ans.} \quad 176 - 105 = 71$