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irakobra [83]
3 years ago
8

It is estimated that 3% of the athletes competing in a large tournament are users of an illegal drug to enhance performance. The

test for this drug is 90% accurate. What is the probability that an athlete who tests positive is actually a user? .3611 .2771 .2177 .2893

Mathematics
1 answer:
yuradex [85]3 years ago
7 0

Answer:

probability that an athlete who tests positive is actually a user

= 0.2177

Step-by-step explanation:

To solve this problem, we create a table that relates the test results and the status of the athletes.

We arbitrarily take a sample of 10000 atheletes to work with whole numbers.

Out of that, 300 are users (3% on the average), and 9700 are non users.

Of the users, 270 users will test positive (90% accuracy) and 30 will test negative.

Similarly, of the non-users, 970 will test positive, and 8730 will test negative.

This gives 1240 positive tests, and 8760 negative tests.

Of the 1240 who test positive, 270 are actually users.

Therefore the probability of the positive test results are actually users

= 270/1240 = 0.2177

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In this attached picture according to the conditions of the problem we have an isosceles trapezoid and since we know that legs are equal (AD=BC=5 cm), we have to calculate bases and height in order to find the area. Working with the triangle BCD, we apply Pythagoras theorem and find that CD = \sqrt{75+25} = 10 cm. Since BDC is a right triangle, applying theorem for the area of triangles, we find that \frac{1}{2} * BF =  \frac{1}{2} * 5 * \sqrt{75} and BF= 0.5\sqrt{75}. Since ABCD is an isosceles trapezoid, triangles ADE and BFC are congruent with Angle Side Angle theorem. Then, DE=FC and with the help of Pythagoras theorem, DE=FC=2.5 cm. Then, AB=EF=5 cm and the area of the trapezoid is  A= BF *  \frac{AB+CD}{2} = 0.5  \sqrt{75}  * \frac{5+10}{2} = 18.75 \sqrt{3}   cm^{2}

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