Of people who died in the United States in recent years, 86% were white, 12% were black, and 2% were Asian. (This ignores a small number of deaths among other races.) Diabetes caused 2.8% of deaths among whites, 4.4% among blacks, and 3.5% among Asians. The probability that a randomly chosen death is a white person who died of diabetes is about:____

Answer:

The probability that a randomly chosen death is a white person who died of diabetes is of 0.0241 = 2.41%.

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: The person is white

Event B: Died of diabeters.

Of people who died in the United States in recent years, 86% were white

This means that [tex]P(A) = 0.86[/tex]

Diabetes caused 2.8% of deaths among whites

This means that [tex]P(B|A) = 0.028[/tex]

The probability that a randomly chosen death is a white person who died of diabetes is about:

This is [tex]P(A \cap B)[/tex]. So

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

[tex]P(A \cap B) = P(B|A)*P(A) = 0.028*0.86 = 0.0241[/tex]

The probability that a randomly chosen death is a white person who died of diabetes is of 0.0241 = 2.41%.

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