Question

An insurance policy on an electrical device pays a benefit of 4000 if the device fails during the first year. The amount of the benefit decreases by 1000 each successive year until it reaches 0. If the device has not failed by the beginning of any given year, the probability of failure during that year is 0.4. What is the expected benefit under this policy

Answer 1

Answer:

Expected benefit under this policy = $ 2694

Step-by-step explanation:

Given - An insurance policy on an electrical device pays a benefit of

            4000 if the device fails during the first year. The amount of the

            benefit decreases by 1000 each successive year until it reaches 0.

            If the device has not failed by the beginning of any given year, the

            probability of failure during that year is 0.4.

To find - What is the expected benefit under this policy ?

Proof -

Let us suppose that,

The benefit = y

Given that, the probability of failure during that year is 0.4

⇒Probability of non-failure = 1 - 0.4 = 0.6

Now,

If the device fail in second year , then

Probability = 0.6×0.4

If the device fail in third year, then

Probability = 0.6×0.6×0.4 = 0.6² × 0.4

Going on like this , we get

If the device is failed in n year, then

Probability = 0.6ⁿ⁻¹ × 0.4

Now,

The probability distribution is-

Benefit , x       4000       3000             2000            1000              0

P(x)                 0.4         0.6×0.4         0.6² × 0.4     0.6³ × 0.4     1 - 0.8704

                      (0.4)       (0.24)            (0.144)         (0.0864)       (0.1296)

At last year, the probability = 1 - (0.4+ 0.24+ 0.144+ 0.0864) = 1 - 0.8704

Now,

We know that,

Expected value ,

E(x) = ∑x p(x)

       = 4000(0.4) + 3000(0.24) + 2000(0.144) + 1000(0.0864) + 0(0.1296)

       = 1600 + 720 + 288 + 86.4 + 0

       = 2694.4

⇒E(x) = 2694.4 ≈ 2694

∴ we get

Expected benefit under this policy = $ 2694