A-LevelMathematicsProbability and statistics9709/62 · Feb/Mar 2022 · Q710 Marks

(a) Two ponds, A and B, each contain a large number of fish. It is known that 2.4% of fish in pond A are carp and 1.8% of fish in pond B are carp. Random samples of 50 fish from pond A and 60 fish from pond B are selected. Use appropriate Poisson approximations to find the following probabilities. (i) The samples contain at least 2 carp from pond A and at least 2 carp from pond B. (ii) The samples contain at least 4 carp altogether. (b) The random variables X and Y have the distributions Po(λ) and Po(μ) respectively. It is given that • P(X = 0) = [P(Y = 0)]², • P(X = 2) = k[P(Y = 1)]², where k is a non-zero constant. Find the value of k.

(a)(i)The samples contain at least 2 carp from pond A and at least 2 carp from pond B.[3]
(a)(ii)The samples contain at least 4 carp altogether.[3]
(b)The random variables X and Y have the distributions Po(λ) and Po(μ) respectively. It is given that • P(X = 0) = [P(Y = 0)]², • P(X = 2) = k[P(Y = 1)]², where k is a non-zero constant. Find the value of k.[4]

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2022). The mark or mark type is shown in brackets.

[B1] 0.024 × 50 [= 1.2] and 0.018 × 60 [= 1.08]
[M1] (1 – e^-1.2(1 + 1.2)) × (1 – e^-1.08(1 + 1.08)) For (1 – e^λ(1+λ)) × (1 − e^μ(1+μ)) any λ, μ (λ ≠μ) Allow one end error on either or both terms
[A1] = 0.0991 (3 sf) Unsupported answer scores maximum SC B1 B1 SC Use of binomial 0.0994 scores B1 only
[M1] λ = 0.024 × 50 +0.018 × 60 or their 1.2 + 1.08 (NB 0.024+0.018 is M0)
[M1] 1-e^-2.28(1+2.28 + 2.28²/2! + 2.28³/3!) any λ and allow one end error
[A1] = 0.197 (3 sf) Unsupported answer scores maximum SC B2
[M1] e^-λ = [e^-μ]² = e^-2μ
[M1] e^-λ × λ²/2! = k[e^-μ × μ]²
[M1] OE. Use of λ = 2μ to find equation in μ and k only (or λ and k only) e^-2μ × 2μ² = k × e^-2μ x μ²
[A1] k=2

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