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

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.

the probability that the insurance company receives four claims in a week is 0.1954

Step-by-step 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

7 ft 6 in means 7 1/2, because 6 in is 1/2 of a foot. 7 ft 8 in means 7 2/3 because 8 in is 2/3 of a foot. 23/3 = 7 2/3. 15/2 = 7 1/2.

23/3   *   15/2   = 57 1/2

step-by-step explanation: