Question

If X has Binomial distribution with parameters n and p such that np =λ, then \(\mathop {\lim }\limits_{n \to \infty } b\left( {x,n,p} \right);x = 0,1,2,.....\) is equal to:
1. \({\frac{{{e^{^{ - \lambda }}}\lambda }}{{x!}}^x}\), x = 0, 1, 2, … 
2. Limit does not exist
3. 0
4. 1

Answer 1

Correct Answer - Option 1 : \({\frac{{{e^{^{ - \lambda }}}\lambda }}{{x!}}^x}\), x = 0, 1, 2, … 

Explanation

Poisson distributionis a limiting case of binomial distribution if it follows conditions

n, the number trials is indefinitely large  that means n tends to infinite

p, the constant probability of success for each trial is indefinitely small p tends to 0

np = λ , is finite so λ/n = p, q = 1 – p

⇒ (1 – λ/n), λ is positive integer

The probability of x successes in a series of n independent trials is

⇒ b(x, n, p) = (n/x)p^xq^{n – x}, x = 0, 1, 2, 3…….n

⇒ b(x, n, p) = (n/x)p^x(1 – p)^{n – x}

∴ (n/x)(p/(1 – p)]^x(1 – p)^{n - x}

p , the constant probability of success for each trial is indefinitely small p tends to 0

np = λ , is finite so λ/n = p, q = 1 – p

⇒ (1 – λ/n), λ is positive integer

The probability of x successes in a series of n independent trials is

⇒ b(x, n, p) = (n/x)p^xq^{n – x}, x = 0, 1, 2, 3…….n

⇒ b(x, n, p) = (n/x)p^x(1 – p)^{n – x}

∴ (n/x)(p/(1 – p)]^x(1 – p)^{n - x}

⇒ [n(n - 1)(n - 2)------(n - x + 1)/x!] × (λ/n)^x/(1 - λ /n)^x[1 - λ/n]ⁿ

⇒ [(1 - 1/n)(1 - 2/n)-----( 1 - (x - 1)/n/x!(1 - λ/n)^x] × λ^x[1 - λ/n]ⁿ

⇒ Lim x → ∞ b(x, n, p) = e^-λ × λ^x/x! ; x = 0, 1, 2, 3, 4 -------,n

Poisson distribution = A random variables X is said to follow poisson distribution if it assumes only non - negative values and its proportionality mass function i s given

by P)X = x) = e-λ × λx/x! where x = 0, 1, 2, 3 ------n and  λ > 0

⇒ p(x, λ) = ∑P(X - x)

⇒ e^-λ∑λ^x/x!

⇒ e^-λ× e^-λ = 1

∴ The corresponding distribution function is F(x) = P(X = x) = ∑P(r) =  e-λ ∑λ²/r!; x = 0, 1, 2 .......