Under an insurance policy, a maximum of five claims may be filed per year by a policyholder. Let pn be the probability that a policyholder files n claims during a given year, where n = 0, 1, 2, 3, 4, 5. An actuary makes the following observations: {p_n \geq p_{n+1}}\,\text{ for }\,{n=0,1,2,3,4} The difference between pn and pn + 1 is the same for n = 0, 1, 2, 3, 4. Exactly 40% of policyholders file fewer than two claims during a given year. Calculate the probability that a random policyholder will file more than three claims during a given year.

Respuesta :

Answer:

0.27

Step-by-step explanation:

We know that p0 + p1 = 2 5 = 0.4

p0 + p1 + p2 + p3 + p4 + p5 = 1.

k = p0 − p1 = p1 − p2 = p2 − p3 = p3 − p4 = p4 − p5 .

Then p1 = p0 − k, p2 = p1 − k = p0 − k − k = p0 − 2k, etc.,

so that pn = p0 − nk for 1 ≤ n ≤ 5.

Thus p0 + ( p0 − k) + ( p0 − 2k) +…+ ( p0 − 5k) = 6 p0 − 15k = 1.

Also 40% = 0.4 = p0 + p1 = p0 + ( p0 − k) = 2 p0 − k.

This results in the following equations:

6 p0 −15k = 1,

2 p0 − k = 0.4,

Solving, the above: k = 1/60 and p0 = 25 /120 .

Therefore: p4 + p5 = ( p0 − 4k) + ( p0 − 5k)

= 17 /120 + 15 /120

= 32 /120 = 0.2667 ~ 0.27

Answer:

0.27

Step-by-step explanation:

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