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Suppose the force of mortality of a person aged $x$ is $\mu_x = \frac{1}{200-x} + \frac{1}{100-x}$ where $x<100$. What is the probability that the person survives at least $t$ more years?

So would this be $$\exp \left(-\int_{x}^{x+t} \frac{1}{200-s} \ ds- \int_{x}^{x+t} \frac{1}{100-s} \ ds \right)$$

$$ = \exp \left(\ln(200-s) |^{x+t}_{x} + \ln(100-s) |^{x+t}_{x} \right)$$

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up vote 4 down vote accepted

The survival function $S(t) = \mathbb{P}(T>t)$, where $T$ is the life-span of the person is defined as $$ S(t) = \exp\left(-\int_0^t \mu_x \mathrm{d}x\right) = \exp\left(-\int_0^t \mu_x \mathrm{d}x\right) = \left(1-\frac{t}{200}\right)\left(1-\frac{t}{100}\right) $$

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