W tAMAT108] Homewor at Albany x 3 c Q www.webassign. net/web/student/Assignment-

Question

W tAMAT108] Homewor at Albany x 3 c Q www.webassign. net/web/student/Assignment-Responses/last?depa15679518#Q10 D My Notes O Ask Your Ti Suppose that a six-sided die is "loaded" so that any particular even numbered face is twice as likely to be observed as any particular odd-numbered face. (a) What are the probabilities of the six simple events? (Hint: Denote these events by o1 o6. Then P(o1) p, Po2) 2p, P(O3) p,..., P(O6)- 2p. Now use a condition on the sum of these probabilities to determine p. Answer as an exact fraction or round your answers to three decimal places.) P(01) P(02) P(O3) PO4) PCOs) POO6) (b) What is the probability that the number showing is an odd number? (Give the answer to three decimal places.) What is the probability that the number showing is at most 3? (Give the answer to three decimal places.) (c) Now suppose that the die is loaded so that the probability of any particular simple event is proportional to the number showing on the corresponding upturned face; that is, P(01) P(02) 2c, Poo6) 6c. What are the probabilities of the six simple events? (Answer as an exact fraction or round your answers to three decimal places.) P(O1) P(02) P(O3) P(04) PCOs)

Explanation / Answer

(a) P(O1)=p=1/9

P(O2)=2p=2/9

P(O3)=p=1/9

P(O4)=2p=2/9

P(O5)=p=1/9

P(O6)=2p=2/9

sum of P(O1)+P(O2)+P(O3)+P(O4)+P(O5)+P(O6)=1

or, 9p=1 or, p=1/9

(b) P(odd number)=1/9 + 1/9 +1/9=3/9=1/3

P(at most 3)=P(O1)+P(O2)+P(O3)=1/9 + 2/9 + 1/9=4/9

(c)

P(O1)=c=1/21

P(O2)=2c=2/21

P(O3)=3c=3/21

P(O4)=4c=4/21

P(O5)=5c=5/21

P(O6)=6c=6/21

sum of P(O1)+P(O2)+P(O3)+P(O4)+P(O5)+P(O6)=1

or, 21c=1 or, c=1/21

P(odd number)=1/21 + 3/21 +5/9=9/21=3/7

P(at most 3)=P(O1)+P(O2)+P(O3)=1/21 + 2/21 + 3/21=6/21=2/7

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