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Lelu [443]
3 years ago
14

From among (a number)

Mathematics
1 answer:
Oksi-84 [34.3K]3 years ago
5 0

para responder hay que ver la pregunta

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On average, Tonya makes $312 more than Emily on a typical day.

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I was just studying that question a few hours ago

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In a large population, 3% of the people are heroin users. A new drug test correctly identifies users 93% of the time and correct
kari74 [83]

Answer:

(a) The probability tree is shown below.

(b) The probability that a person who does not use heroin in this population tests positive is 0.10.

(c) The probability that a randomly chosen person from this population is a heroin user and tests positive is 0.0279.

(d) The probability that a randomly chosen person from this population tests positive is 0.1249.

(e) The probability that a person is heroin user given that he/she was tested positive is 0.2234.

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = a person is a heroin user

<em>Y</em> = the test is correct.

Given:

P (X) = 0.03

P (Y|X) = 0.93

P (Y|X') = 0.99

(a)

The probability tree is shown below.

(b)

Compute the probability that a person who does not use heroin in this population tests positive as follows:

The event is denoted as (Y' | X').

Consider the tree diagram.

The value of P (Y' | X') is 0.10.

Thus, the probability that a person who does not use heroin in this population tests positive is 0.10.

(c)

Compute the probability that a randomly chosen person from this population is a heroin user and tests positive as follows:

P(X\cap Y)=P(Y|X)P(X)=0.93\times0.03=0.0279

Thus, the probability that a randomly chosen person from this population is a heroin user and tests positive is 0.0279.

(d)

Compute the probability that a randomly chosen person from this population tests positive as follows:

P (Positive) = P (Y|X)P(X) + P (Y'|X')P(X')

                  =(0.93\times0.03)+(0.10\times0.97)\\=0.1249

Thus, the probability that a randomly chosen person from this population tests positive is 0.1249.

(e)

Compute the probability that a person is heroin user given that he/she was tested positive as follows:

P(X|positive)=\frac{P(Y|X)P(X)}{P(positive)} =\frac{0.93\times0.03}{0.1249}= 0.2234

Thus, the probability that a person is heroin user given that he/she was tested positive is 0.2234.

6 0
3 years ago
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