Consider three elements {e1; e2; e3} put intorandom order as in the table below.
ID: 2952187 • Letter: C
Question
Consider three elements {e1; e2; e3} put intorandom order as in the table below.list ¦e1e2e3 ¦e1e3e2 ¦e2e1e3 ¦e2e3e1¦e3e1e2¦ e3e2e1
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probability ¦ 1/2 ¦ 1/4 ¦ 1/8 ¦ 1/16 ¦ 1/32¦ 1/32
Calculate the following quantities:
(a) P(e3 precedes e2); (b) P(e2 is in position 2); (c) E( positionof e1)
Consider three elements {e1; e2; e3} put intorandom order as in the table below.
list ¦e1e2e3 ¦e1e3e2 ¦e2e1e3 ¦e2e3e1¦e3e1e2¦ e3e2e1
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probability ¦ 1/2 ¦ 1/4 ¦ 1/8 ¦ 1/16 ¦ 1/32¦ 1/32
Calculate the following quantities:
(a) P(e3 precedes e2); (b) P(e2 is in position 2); (c) E( positionof e1)
Consider three elements {e1; e2; e3} put intorandom order as in the table below.
list ¦e1e2e3 ¦e1e3e2 ¦e2e1e3 ¦e2e3e1¦e3e1e2¦ e3e2e1
________________________________________________________
probability ¦ 1/2 ¦ 1/4 ¦ 1/8 ¦ 1/16 ¦ 1/32¦ 1/32
Calculate the following quantities:
(a) P(e3 precedes e2); (b) P(e2 is in position 2); (c) E( positionof e1)
Explanation / Answer
(a) P(e3 precedes e2) We can see from the table that e3 precedes e2 in (e1e3e2), (e3e1e2), and (e3e2e1) which have the respectiveprobabilities of (1/4), (1/32), and (1/32). hence P(e3 precedes e2)= P(e1e3e2)+P(e3e1e2)+P(e3e2e1) =(1/4)+(1/32)+(1/32) = (5/16) = .3125 (b) P(e2 is in position 2) We can see from the table that e2 is in position 2 in (e1e2e3) and(e3e2e1) which have the respective probabilities (1/2) and(1/32). so, P(e2 is in position 2) = P(e1e2e3)+P(e3e2e1) =(1/2)+(1/32) = (17/32) = .53125 (c) E(position of e1) e1 in 1stposition e1 in 2nd position e1in 3rd position P (1/2)+(1/4)=3/4 (1/8)+(1/32)=5/32 (1/16)+(1/32) =3/32