Three rotten apples are mixed accidently with seven good apples and four apples are drawn one by one without replacement. Let the random variable denote the number of rotten apples. If and represent mean and variance of , respectively, then is equal to
- A.25
- B.250
- C.20
- D.30
Show answer and solution
Correct option: C
Solution
X follows Hypergeometric distribution: drawing 4 from 10 (3 rotten, 7 good).
P(X = k) = C(3,k)·C(7,4-k)/C(10,4), for k = 0,1,2,3.
C(10,4) = 210.
P(X=0) = C(3,0)·C(7,4)/210 = 35/210 = 1/6
P(X=1) = C(3,1)·C(7,3)/210 = 3·35/210 = 105/210 = 1/2
P(X=2) = C(3,2)·C(7,2)/210 = 3·21/210 = 63/210 = 3/10
P(X=3) = C(3,3)·C(7,1)/210 = 7/210 = 1/30
For Hypergeometric: μ = n·K/N = 4·3/10 = 6/5
σ² = n·K·(N-K)·(N-n) / [N²·(N-1)] = 4·3·7·6 / [100·9] = 504/900 = 14/25
Now: μ² = 36/25, σ² = 14/25
μ² + σ² = 50/25 = 2
10(μ² + σ²) = 10·2 = 20
P(X = k) = C(3,k)·C(7,4-k)/C(10,4), for k = 0,1,2,3.
C(10,4) = 210.
P(X=0) = C(3,0)·C(7,4)/210 = 35/210 = 1/6
P(X=1) = C(3,1)·C(7,3)/210 = 3·35/210 = 105/210 = 1/2
P(X=2) = C(3,2)·C(7,2)/210 = 3·21/210 = 63/210 = 3/10
P(X=3) = C(3,3)·C(7,1)/210 = 7/210 = 1/30
For Hypergeometric: μ = n·K/N = 4·3/10 = 6/5
σ² = n·K·(N-K)·(N-n) / [N²·(N-1)] = 4·3·7·6 / [100·9] = 504/900 = 14/25
Now: μ² = 36/25, σ² = 14/25
μ² + σ² = 50/25 = 2
10(μ² + σ²) = 10·2 = 20
[JEE Main 2023 · 29 Jan · Shift 1]