The random variable A has the distribution Po(1.5). A₁ and A₂ are independent values of A.
- (a)Find P(A₁ + A₂ < 2).[3]
- (b)Given that A₁ + A₂ < 2, find P(A₁ = 1).[4]
- (c)Give a reason why A₁ – A₂ cannot have a Poisson distribution.[1]
Mark scheme
Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.
- [B1] (a) λ = 3
- [M1] (a) e⁻³(1+3)
- [A1] (a) = 0.199 (3 sf)
- [M1] (b) P(A₁ = 1 and A₁ + A₂ < 2) = P(A₁ = 1) × P(A₂ = 0)
- [A1] (b) e⁻¹⁵ × 1.5 × e⁻¹⁵ = 0.0747
- [M1] (b) P(A₁=1 | A₁+A₂<2) = P(A₁=1 and A₁ + A₂ < 2) / P(A₁ + A₂ < 2) = ‘0.0747’ / ‘0.199’
- [A1] (b) 3/8 or 0.375 (3 sf)
- [B1] (c) Takes negative values
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