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Probability of Incorrect Diagnosis among Employees

Calculate the probability of at least one employee being incorrectly diagnosed with a disease when they do not have it.

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Probability of Incorrect Diagnosis among Employees

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  1. WARM – UP The diagnosis for having a certain disease has a 0.002 probability of being wrong (False Positive). A company has 520 employees that must undergo a test to determine if they have this disease. What is the probability that at least one of them is incorrectly diagnosed as having the disease when he or she does not have it? P(At least one false positive) = 1 – P(No Incorrect Results) = 1 – P(520 Healthy = Neg.) = 1 – 0.998520 = 0.6469

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  3. A B Ac and Bc CHAPTER 15 General Addition Rule for Probability (OR) P(A U B) = P(A) + P(B) – P(A ∩ B) The Probability that Neither Event A nor B will occur is: P(Ac∩ Bc) = 1 – P(A U B)

  4. EXAMPLE: Let event X = Ticket for Speeding. Let event Y = Ticket for Running a Stop Sign. If P(X) = 0.12, P(Y) = 0.09, and P(X ∩ Y) = 0.07 find: 1.) Find the probability that you will get a ticket for speeding OR a ticket for running a stop sign. 2.) Find the probability that you will NOT get either Ticket. P(A U B) = P(A) + P(B) – P(A ∩ B) P(X U Y) = 0.12 + 0.09 – 0.07 0.14 P(Ac∩ Bc) = 1 – P(A U B) P(Xc ∩ Yc) = 1 – 0.14 0.86

  5. EXAMPLE: If P(A) = 0.40, P(B) = 0.20, and P(A U B) = 0.58 1. Find P(A ∩ B)=? 2. Are events A and B Independent, Mutually Exclusive, or Neither? P(A U B) = P(A) + P(B) – P(A ∩ B) 0.58 = 0.40 + 0.20 – P(A ∩ B) 0.58 = 0.60 – P(A ∩ B) -0.02 = – P(A ∩ B) 0.02 = P(A ∩ B) Independent means P(A ∩ B) = P(A) P(B) ? 0.02 = 0.40 • 0.20 =.08 Mutually Exclusive means P(A ∩ B) = 0 A and B are neither Independent or Mutually Exclusive

  6. EXAMPLE: Employment data at a large company reveal that 72% of the workers are married, 44% are college graduates, and half of the college graduates are married. What is the probability that a randomly chosen worker: a.) is married or a college graduate? b.) is married but not a college graduate? c.) is neither married nor a college graduate? P(M U G) = P(M) + P(G) – P(M ∩ G) = 0.72 + 0.44 – 0.22= 0.94 P(M ∩ Gc) = P(M) – P(M ∩ G) = 0.72 – 0.22= 0.50 P(Mc∩ Gc) = 1 – P(M U G) = 1 – [P(M) + P(G) – P(M ∩ G)] = 1 – [0.72 + 0.44 – 0.22] = 1 – [0.94]= 0.06

  7. EXAMPLE: Democrat Republican NO Party West 39 17 12 Northeast 15 30 12 Southeast 30 31 16 68 57 77 84 78 40 202 P(D) = 0.4158 • Find the Probability of being a Democrat. • Find the Probability of living in the Northeast? • From the Chart, find the Probability of being a Democrat AND living in the Northeast. • Is living in the Northeast independent from being a Democrat? Why or Why not? P(N) = 0.2822 P(D ∩ N) = 0.0743 NO, b/c P(D ∩ N) ≠ P(D)·P(N)

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  9. The American Red Cross says that about 45% of the US Population has Type O blood, 40% Type A, 11% Type B, and the rest Type AB. a.) Selecting ONE individual, what is the probability that: 1. has Type AB blood? 2. has Type A or Type B? 3. is NOT Type O? b.) Among four potential donors, what is the probability that: 1. all are Type O? 2. no one is Type AB? 3. they are not all Type A? 4. at least one person is Type B? • Page 341 #24 P(AB) = 1 – P(OUAUB) = 0.04 P(AUB) = 0.51 P(OC) = 0.55 P(O∩O∩O∩O) = 0.041 P(ABC∩ABC∩ABC∩ABC) = 0.849 1 – P(A∩A∩A∩A) = 0.974 0.373 1 – P(BC∩BC∩BC∩BC) =

  10. The Cross Country Team says that about 45% its members are Seniors, 30% Junior, 18% Sophomores, and the rest Freshman. 1.) Selecting ONE individual, what is the probability that he: a. is a Freshman? b. is a Senior or Junior? c. is NOT a Senior? 2.) Among four potential runners, what is the probability that: a. all are Seniors? b. no one is Freshman? c. they are not all Juniors? d. at least one person is Sophomore? Warm - Up P(F) = 1 – P(SeUJUSo) = 0.07 P(SeUJ) = 0.75 P(SeC) = 0.55 P(Se∩Se∩Se∩Se) = 0.041 0.748 P(FC∩FC∩FC∩FC) = 1 – P(J∩J∩J∩J) = 0.9919 0.548 1 – P(SoC∩SoC∩SoC∩SoC) =

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