1. Given Equations:
Binomial distribution ke liye hum jaante hain:
Pehli condition ke anusar:
Kyuki 1−q=p, isliye:
2. Second Condition ka use karke n aur p nikalna:
Hume diya gaya hai 2P(X=2)=3P(X=1):
2⋅(2n)p2qn−2=3⋅(1n)p1qn−1
2⋅2n(n−1)⋅p2qn−2=3⋅n⋅p⋅qn−1
3. Equations ko solve karna:
Eq (i) se hum likh sakte hain n=p21. Isse Eq (ii) mein rakhne par:
Solving this quadratic: (2p−1)(p−1)=0.
Yahan p=1/2 (kyuki p=1 lene par variance 0 ho jayega).
Agar p=1/2, toh Eq (i) se: n(1/2)2=1⟹n=4.
4. Final Calculation:
Hume nikalna hai n^2 P(X > 1):
P(X > 1) = 1 - [P(X=0) + P(X=1)]
P(X=0)=(04)(1/2)0(1/2)4=1/16
P(X=1)=(14)(1/2)1(1/2)3=4/16
P(X > 1) = 1 - \left(\frac{1}{16} + \frac{4}{16}\right) = 1 - \frac{5}{16} = \frac{11}{16}
Ab, n^2 P(X > 1) = 4^2 \cdot \frac{11}{16} = 16 \cdot \frac{11}{16} = \mathbf{11}.
