7. 10 points] In a particular college statistics course, twenty percent of the s

Question

7. 10 points] In a particular college statistics course, twenty percent of the students had previously tak en a statistics course during high school while the others had not. Among those students who had taken a high school course, twenty-five percent received an A in their college course; while only fifteen percent of those students with no prior exposure received an A in the college course. Suppose that one student from the college course is selected at random (a) Find the probability that they had taken statistics during high school and receive an A in their college course. 3 points] (b) Find the probability that they receive an A in the college course. [3 points] (c) Suppose that you overhear the student boasting about receiving an A in their college statistics course. Find the probability that they had not taken a statistics course during high school. 4 points] Section 2.5 12 points] Suppose that the proportions of blood pheno types in a particular population are as given in the table below. Suppose, further, that the phenotypes of two randomly selected individuals are independent of each another. 3 points each] 8 B O Proportion 0.40 0.1 0.04 0.45 not (a) Find the probability that both phenotypes are A. b) Find the probability that neither phenotype is AB (c) Find the probability that the phenotypes of the two individuals match. (d) Suppose that three individuals are selected independently of each other. Find the probability that at least one of the three has Type O blood.

Explanation / Answer

3 b) probability =P(A|B) =P(AnB)/P(B) =(P(A)+P(B)-P(AUB))/P(B) =(0.40+0.55-0.63)/0.55=0.5818

7c)

probability of receiveing A =P( had stat and got and A +had not have stat and got an A)

=0.2*0.25+(1-0.2)*0.15=0.17

therefore probability of not having stat given got an A =P(had not have stat and got an A)/P(A)=(1-0.2)*0.15/0.17

=0.7059

8 c)

probability that phenotype matches =P( both are A +both are B +both are AB +both are O)

=0.40*0.4+0.11*0.11+0.04*0.04+0.45*0.45=0.3762

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