From a lot of 12 items containing 3 defectives, a sample of 5 items is drawn at random. Let the random variable 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 is w

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Solution

Q.
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 $g c d (m, n) = 1,$ then $n - m$ is equal to ______ . [2024]

Ans.
(71)

X 0 1 2 3

P(X) $\frac{{}^{9}{C_{5}}}{{}^{12}{C_{5}}}$ $\frac{{}^{9}{C_{4} \times {}^{3}{C_{1}}}}{{}^{12}{C_{5}}}$ $\frac{{}^{9}{C_{3} \times {}^{3}{C_{2}}}}{{}^{12}{C_{5}}}$ $\frac{{}^{9}{C_{2} \times {}^{3}{C_{3}}}}{{}^{12}{C_{5}}}$

$Mean = \sum X P (X) = 0 + \frac{126 \times 3}{792} + \frac{2 \times 84 \times 3}{792} + \frac{3 \times 36 \times 1}{792}$

$= \frac{378 + 504 + 108}{792} = \frac{990}{792} = \frac{55}{44}$

$Variance = \sum X^{2} P (X) - (\sum X P {(X))}^{2}$

$= \frac{95}{44} - {(\frac{55}{44})}^{2} = \frac{4180 - 3025}{1936} = \frac{1155}{1936} = \frac{105}{176} = \frac{m}{n}$

$So, n - m = 176 - 105 = 71$

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