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 variance of X is m/n, where gcd(m, n)=1, then n-m is equal to _________.

Solution

Maths sir · 3 steps

Without replacement means hypergeometric: tabulate the probabilities, then use E[X²]-(E[X])².

1. Set up X. There are 3 defective and 9 good items, and 5 are drawn, so X can be 0, 1, 2 or 3. The order of drawing does not matter, so P(X=k) = C(3, k)C(9, 5-k)/C(12, 5), with C(12, 5) = 792.

   k	Ways C(3, k)C(9, 5-k)	k × ways	k² × ways
   0	1 × 126 = 126	0	0
   1	3 × 126 = 378	378	378
   2	3 × 84 = 252	504	1008
   3	1 × 36 = 36	108	324
   Total	792	990	1710

2. Mean and square. E[X] = 990/792 = 5/4 and E[X²] = 1710/792 = 95/44.

   Var(X) = 95/44 - 25/16 = (380 - 275)/176 = 105/176

3. So. 105 = 3 × 5 × 7 and 176 = 2⁴ × 11 share no factor, so m = 105, n = 176 and n - m = 71.

The common slip: 1. Using the binomial formula 5 × 1/4 × 3/4 = 15/16 ignores the without-replacement factor 7/11 and gives 16 - 15 = 1.

The idea to keep. Drawing without replacement is not binomial. The variance is np(1-p)×(N-n)/(N-1), and that extra factor makes it smaller than the binomial value.

Answer from the NTA final answer key. Solution by PrepWiser.