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
.
