1. The following table lists the joint probabilities associated with smoking and

Question

1. The following table lists the joint probabilities associated with smoking and lung disease among 60-to-65 year- old men. One 60-to-65 year old man is selected at random. What is the probability of the following events? Smoker Nonsmoker Has Lung Disease 0.1 No Lung Disease 0.2 0.04 0.66 A. He is a smoker B. He does not have lung disease: C. He has lung disease given that he is a smoker: D. He has lung disease given that he does not smoke 2. Suppose that A and B are two independent events for which P(A) = 0.31 and P(B) = 0.76. Find each of the following: A. PAB)- B. RBA) C. P(A and B)_ D. P(A or B)_ 3. A firm has classified its customers in two ways: according to whether the account is overdue and whether the

Explanation / Answer

1)

a)

P(Smoker) = (0.1 + 0.2) = 0.3

b)

P(No lung disease) = 0.2 + 0.66 = 0.86

c)

P(Has lung disease | Smoker) = P(Has lung disease and Smoker) / P(Smoker)

= 0.1 / 0.3

= 0.3333

d)

P(Has lung disease | nonsmoker) = P(Has lung disease and nonsmoker) / P(nonsmoker)

= 0.04 / 0.70

= 0.0571

2)

The events A and B are independents means P( A and B) = P(A) * P(B)

Given, P(A) = 0.31 , P(B) = 0.76

a)

P( A | B) = P( A and B) / P(B)

= P(A) * P(B) / P(B)

= P(A)

= 0.31

b)

P(B | A) = P( A and B) / P(A)

= P(A) * P(B) / P(A)

= P(B)

= 0.76

c)

P(A and B) = P(A) * P(B)

= 0.31 * 0.76

= 0.2356

d)

P( A or B) = P(A) + P(B) - P(A and B)

= 0.31 + 0.76 - 0.2356

= 0.8344

Smoker Nonsmoker Total has lung disease 0.1 0.04 0.14 No lung disease 0.2 0.66 0.86 Total 0.3 0.70 1.00

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.