6. I7 points DevoreStat9 2.E.045. My Notes Ask Your Teacher The population of a

Question

6. I7 points DevoreStat9 2.E.045. My Notes Ask Your Teacher The population of a particular country consists of three ethnic groups. Each individual belongs to one of the four major blood groups. The accompanying joint probabity table gives the proportions of individuals in the various ethnic group-blood group combinations. Blood Group 0 AB 1 0.082 0,111 0.010 0,004 Ethnic Group 2 0.130 0,141 0.018 0.004 3 0.215 0.207 0.058 0.020 suppose that an individual is randomly selected from the population, and define events by A (a) Calculate PA), P(C), and PIA n C). (Enter your answers to three decimal places.) {type A selected}, B {type B selected), and C = {ethnic group 3 selected? P(A)- (b) Calculate both PAI C) and PC I A). (Round your answers to three decimal places.) PCA IC)- P(C | A)= Explain in context what each of these probabilities represents. (Select all that apply.) If we know that the individual came from ethnic group 3, the probability that he has type A is given by A). 1f a person has type B blood, the probability that he is from ethnic group 3 is given by p(C | A) If a person has type A blood, the probability that he is from ethnic group 3 is given by A) If a person has type B blood, the probability that he is from ethnic group 3 is given by P(A | C) If we know that the individual came from ethnic group 3, the probability that he has type A is given by PA 1 C). If a person has type A blood, the probability that he is from ethnic group 3 is given by P(A 1 C) (c) If the selected individual does not have type B blood, what is the probability that he or she is from ethnic group 1? (Round your answer to three decimal places.)

Explanation / Answer

Ans:

For,Marginal probabilties(add row and columns)

a)P(A)=0.459

P(C)=0.5

P(A and C)=0.207

b)

P(A/C)=P(A and C)/P(C)=0.207/0.5=0.414

P(C/A)=P(A and C)/P(A)=0.207/0.459=0.451

If a person has type A blood,the probability that he s from ethnic group 3 is given by P(C/A)

If we know that individual came from group 3 ,the probability that he has type A is given by P(A/C)

(3rd and 5th are correct)

c)

P(not have type B blood)=1-0.086=0.914

P(group 1 /not B)=P(group 1 and not B)/P(not B)=(0.082+0.111+0.004)/0.914=0.216

O A B AB Marginal 1 0.082 0.111 0.01 0.004 0.207 2 0.13 0.141 0.018 0.004 0.293 3 0.215 0.207 0.058 0.02 0.5 Marginal 0.427 0.459 0.086 0.028 1

Related Questions

Navigate

• Browse (All)
• Browse 6
• Subjects

---

---

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