Name: Major: ns 11 and 12 are based on the following Suppose there is a bacteria

Question

Name: Major: ns 11 and 12 are based on the following Suppose there is a bacteria present in humans. The probability any human has this bacteria is 0.40. There exists a test for the bacteria, and although the test is good, isn't erfect. If a person has the bacteria, the probability that the test will say they have the bacteria is 0.90. If a perso the bacteria the probability that the test will say they do not have the bacteria is 0.91. Strongly suggest you define each event in this problem BEFORE attempting any computations * 11. (3 pts) No partial credit. From the problem data, what is the probability that a randomly selected human does not have the bacteria? Circle the correct answer 0.10 0.40 (a60) 090 0.09 0.91 12. (3 pts) No partial credit. If a randomly selected human's test says they do not have the bacteria, what is the probability that they do not have the bacteria? Circle the correct answer 0.9057 0.9179 0.9317 0.9552 0.9822 0.9982 2 s) No partial credit. De termine the numerical value of k that makes p(x) a valid probability mass functio Express your answer to 4 decimal places. Place your answer on this line: 2.

Explanation / Answer

Preparatory Work for Q11 and Q12

Let A represent the event that a human has bacteria; B represent the event that the test says the human has bacteria. Then, trivially, ACBC respectively represent the events that a human does not have bacteria and the test says the human does not have bacteria.

With this notation, the given data would be:

P(A) = 0.4 ………………………………………………………………………………(1)

And hence P(AC) = 0.6 ……………………………………………………………………(2)

P(B/A) = 0.9………………………………………………………………………………(3)

And hence P(BC/A) = 0.1 ……………………………………………………………………(4)

P(BC/AC) = 0.91 ……………………………………………………………………(5)

P(B/AC) = 0.09 ……………………………………………………………………(6)

Back-up Theory

If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then

Conditional Probability of B given A, denoted by P(B/A) = P(B A)/P(A)..….(7)

P(B) = {P(B/A) x P(A)} + {P(B/AC) x P(AC)}………………………………….(8)

P(A/B) = P(B/A) x { P(A)/P(B)}……………………………..………………….(9)

Now, to work out the solution,

Q11

Probability a randomly selected human does not have bacteria = P(AC)

= 0.6 ANSWER [vide (2) above]

Q12

Probability a randomly selected human does not have bacteria if the test says the human does not have bacteria = P(AC/BC)

= P(BC/AC) x P(AC)/ P(BC) [vide (9) above]

= (0.91 x 0.6)/P(BC) [vide (5) and (2) above]

Now, [vide (8) above],

P(BC) = {P(BC/A) x P(A)} + {P(BC/AC) x P(AC)}

= (0.1 x 0.4) + (0.91 x 0.6) [vide (5), (6), (4) and (1) above]

= 0.586 ANSWER

Q13

Unable to address because the function p(x) is cropped in the image posted.

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