This question is incomplete, the complete question is;
An insurance company determines that the number of claims received in a week follows a Poisson distribution. The probability that the insurance company receives at most one claim in a week is 5/13 times the probability that the insurance company receives at most two claims in a week.
Calculate the probability that the insurance company receives four claims in a week.
Answer:
the probability that the insurance company receives four claims in a week is 0.1954
Explanation:
P(at most one claim) = P(X ≤ 1)
and
P(at most two claims) = P(X ≤ 2)
for P(X ≤ 1) = (5/13) × P(X ≤ 2)
P(X ≤ 1 ) = (5/13) × (P(X ≤ 1) + P(X = 2))
(8/13) × P(X ≤ 1) = (5/13) × P(X = 2)
8 × ( e^(-λ) λ⁰ / 0! + e^(-λ)λ₁/1! ) = 5 × (e^(-λ)λ²/2!
16( 1 + λ ) = 5 × λ²
5 × λ²- 16 × λ - 16 = 0
λ = 4
Therefore the probability that the insurance company receives four claims in a week is P(X = 4) =e⁻⁴4⁴/4! = 0.1954