Conditional Probability and life expectancy

Question

In a population of 100,000 females, 89.835% can expect to live to age 60, while 57.062% can expect to live to age 80. Given that a woman is 60, what is the probability that she lives to age 80?

Using conditional probability Pr(A given B) I am coming up with an answer of .57062. I was wondering if anyone could tell me if my understanding is correct.

Answer 1

Let $ S $ denote the event that the woman lives up to age $ 60 $. Let $ E $ denote the event that she lives upto age $ 80 $. We need to find $ P(E | S) $.

Now, by bayes rule,

$$ P( E | S ) = \dfrac{P(S | E) \times P(E)}{P(S)} $$

Now, $ P(S|E) = 1$ since every person who lives till $ 80 $ lives till $ 60 $.

Therefore,

$$ P(E|S) = \dfrac{1.0 \times 0.57062}{0.89835} \approx 0.635186 \approx 63.51 \% $$

Answer 2

Since, in this population of 100,000 women, "89.835% can expect to live to age 60, while 57.062% can expect to live to age 80", 0.89825*100000= 89835 can expect to live to age 60 and 0.57062*100000= 57062 can expect to live to age 80. Of course, those who lived to age 80 must have been among those who lived to 60 so 57062/89835= 0.63518 or 63.518% of those who lived to be 60 can expect to live to be 80. (adijo, I would have rounded 63.518% to 63.52% rather than 63.51% but the data was given to three decimal places rather than 2.)