Question

Suppose that in a certain country, 10% of the elderly people have diabetes (call this Event D). It is also known that 30% of the elderly people are living below poverty level (call this Event P), 35% of the elderly population falls into at least one of these categories, and 5% of the elderly population falls into both of these categories. If a randomly selected elderly person is living below the poverty level, what is the probability that she/he has diabetes

Answer 1

0.1667 = 16.67% probability that she/he has diabetes

Explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

`P(B|A) = (P(A \cap B))/(P(A))`

In which

P(B|A) is the probability of event B happening, given that A happened.

`P(A \cap B)` is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Living below the poverty level.

Event B: Having diabetes.

30% of the elderly people are living below poverty level.

This means that
`P(A) = 0.3`

5% of the elderly population falls into both of these categories.

This means that
`P(A \cap B) = 0.05`

If a randomly selected elderly person is living below the poverty level, what is the probability that she/he has diabetes

`P(B|A) = (P(A \cap B))/(P(A)) = (0.05)/(0.3) = 0.1667`

0.1667 = 16.67% probability that she/he has diabetes