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Sagot :
Answer:
0.1667 = 16.67% probability that she/he has diabetes
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] 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 [tex]P(A) = 0.3[/tex]
5% of the elderly population falls into both of these categories.
This means that [tex]P(A \cap B) = 0.05[/tex]
If a randomly selected elderly person is living below the poverty level, what is the probability that she/he has diabetes
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.05}{0.3} = 0.1667[/tex]
0.1667 = 16.67% probability that she/he has diabetes
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