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Sagot :
Answer:
a) 0.3064 = 30.64% probability that athlete will test positive for prohibited drug use.
b) 0.2693 = 26.93% probability that this drug user never used the prohibited drug.
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.
A. If an athlete is selected at random, what is the probability that athlete will test positive for prohibited drug use?
100 - 5 = 95% of 8%
100 - 13 = 87% of 17%
11% of 75%
So
[tex]p = 0.95*0.08 + 0.87*0.17 + 0.11*0.75 = 0.3064[/tex]
0.3064 = 30.64% probability that athlete will test positive for prohibited drug use.
B. If an athlete tests positive after a drug test, what is the probability that this drug user never used the prohibited drug?
Conditional Probability
Event A: Tests Positive
Event B: Never used the drug.
0.3064 = 30.64% probability that athlete will test positive for prohibited drug use
This means that [tex]P(A) = 0.3064[/tex]
Probability of testing positive while never using the drug.
11% of 75%. So
[tex]P(A \cap B) = 0.11*0.75 = 0.0825[/tex]
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0825}{0.3064} = 0.2693[/tex]
0.2693 = 26.93% probability that this drug user never used the prohibited drug.
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