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

i would say c is the best answer, hope it's right!

good luck! =)

cutie~

7. d. 64

8. i. 32

9. c. 125

10. j. none of the above

11. d. 81

12. i. 216

13. e. none of the above

14. g. 240

15. e. none of the above

16. g. 32

17.   e. none of the above