Question

If X has Poisson distribution with parameter m and P[X = 2] = P[X = 3], then find P[X ≥ 2].

[Given: e⁻³ = 0.0497]

Solution: X ∼ P(m)

∴ P[X = x] = `(e^(-m)m^x)/(x !)`

∴ P[X = 2] = `(e^(-m)m^2)/(2 !)`

∴ P[X = 3] = `square`

Now P[X = 2] = P[X = 3]

∴ `(e^(-m)m^2)/(2 !) = (e^(-m)m^3)/(3 !)`

∴ m = `square`

Now P[X ≥ 2] = 1 − `square`

= 1 − [P(X = 0) + P(X = 1)]

= `1 - [(e^(-3)3^0)/(0!) + (e^(-3)3^1)/(1!)]`

= 1 − e⁻³[1 + 3]

= 1 − 0.0497 × 4

Hence P[X ≥ 2] is `square`.

Answer 1

X ∼ P(m)

∴ P[X = x] = `(e^(-m)m^x)/(x !)`

∴ P[X = 2] = `(e^(-m)m^2)/(2 !)`

∴ P[X = 3] = \[\boxed{\frac{e^{-m}m^3}{3 !}}\]

Now P[X = 2] = P[X = 3]

∴ `(e^(-m)m^2)/(2 !) = (e^(-m)m^3)/(3 !)`

∴ m = \[\boxed{3}\]

Now P[X ≥ 2] = 1 − \[\boxed{P(X < 2)}\]

= 1 − [P(X = 0) + P(X = 1)]

= `1 - [(e^(-3)3^0)/(0!) + (e^(-3)3^1)/(1!)]`

= 1 − e⁻³[1 + 3]

= 1 − 0.0497 × 4

Hence P[X ≥ 2] is \[\boxed{0.8012}\].