3. Solve the following below that involves probability: A Seniors at a certain h

Question

3. Solve the following below that involves probability:

A

Seniors at a certain high school took a survey regarding future plans. All plan to attend college some time; however, 77% plan to go to college immediately following high school. Of those who plan to attend college immediately following high school, 18% plan to major in Math. Of those who do not plan to attend college immediately following high school, 11% plan to major in Math. What is the probability that a randomly chosen senior does not plan to attend college immediately following high school and plans to major in Math.

a) 0.1386

b) 0.1800

c) 0.8900

d) 0.0253

e) 0.1100

f) None of the above.

B

A new test to detect TB has been designed. It is estimated that 87% of people taking this test have the disease. The test detects the disease in 97% of those who have the disease. The test does not detect the disease in 99% of those who do not have the disease. If a person taking the test is chosen at random, what is the probability of the test indicating that the person does not have the disease?

a) 0.0261

b) 0.0100

c) 0.1548

d) 0.0300

e) 0.9900

f) None of the above.

C.

Urn A contains 4 yellow balls and 10 red balls. Urn B contains 12 yellow balls and 11 red balls. Urn C contains 3 yellow balls and 8 red balls. An urn is picked randomly (assume that each urn is equally likely to be chosen), and then a ball is picked from the selected urn. What is the probability that the chosen ball came from urn B, given that it was a yellow ball?

a) 0.2645

b) 0.0400

c) 0.0569

d) 0.4830

e) 0.2525

f) None of the above.

Explanation / Answer

Question A)

Here, we are given that:

P( college ) = 0.77, therefore P( no college ) = 1 - 0.77 = 0.23

Also, we are given that:

P( math | college ) = 0.18 and P( math | no college ) = 0.11

Therefore the required probability here is computed using bayes theorem as:

P( no colege and math ) = P( math | no college )P( no college ) = 0.11*0.23 = 0.0253

Therefore b) 0.0253 is the required probability here.

B) P( disease ) = 0.87, therefore P( no disease ) = 0.13

P( positive | disease ) = 0.97, therefore P( negative | disease ) = 0.03

P( negative | no disease ) = 0.99, therefore P( positive | no disease ) = 0.01

P( negative ) = P( negative | no disease ) P( no disease ) + P( negative | disease ) P( disease )

P( negative ) = 0.99*0.13 + 0.03*0.87 = 0.1548

Therefore C) 0.1548 is the required probability here.

(C) P( yellow | A) = 4/14

P( yellow | B) = 12/23

P( yellow | C) = 3/11

Therefore, using law of total probability, we get:

P( yellow ) = (1/3)*(4/14 + 12/23 + 3/11 ) = 0.36

Now given that a yellow ball is drawn, probability that is was drawn from urb B is computed as:

P( B | yellow ) = P( yellow | B)P(B) / P( yellow )

P( B | yellow ) = (12/23)*(1/3) / 0.36

P( B | yellow ) = 0.4830

Therefore d) 0.4830 is the required probability here.

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