**5. Given the probability model shown in table 4-9, the real value label of A1A1A3A2A3A1 is obtained.**

**Table 4-9 The probabilistic model of Exercise 5, exercise 6** |

Letter Probability |

A1 0.2 |

A2 0.3 |

A3 0.5 |

Solution:

P (A1) =0.2, P (A2) =0.3, P (A3) =0.5

FX (0) =0,FX (1) =0.2, FX (2) =0.5, FX (3) =1.0, U (0) =1, L (0) =0

by X (AI) =i x (A1) =1,x (A2) =2,x (A3) =3

by formula, L (n) =l (n-1) + (U (n-1)-L (n-1)) Fx (xn-1)

U (n) =l (n-1) + (U (n-1)-L (n-1)) Fx (xn)

When the A1 first appears:

L (1) =l (0) + (U (0)-L (0)) Fx (0) =0

U (1) =l (0) + (U (0)-L (0)) Fx (1) =0.2

When A1 appears for the second time:

L (2) =l (1) + (U (1)-L (1)) Fx (0) =0

U (2) =l (1) + (U (1)-L (1)) Fx (1) =0.04

When A3 appears for the third time:

L (3) =l (2) + (U (2)-L (2)) Fx (2) =0.02

U (3) =l (2) + (U (2)-L (2)) Fx (3) =0.04

Fourth time when A2 appears:

L (4) =l (3) + (U (3)-L (3)) Fx (1) =0.024

U (4) =l (3) + (U (3)-L (3)) Fx (2) =0.03

Fifth time when A3 appears:

L (5) =l (4) + (U (4)-L (4)) Fx (2) =0.027

U (5) =l (4) + (U (4)-L (4)) Fx (3) =0.03

Sixth time when A1 appears:

L (6) =l (5) + (U (5)-L (5)) Fx (0) =0.027

U (6) =l (5) + (U (5)-L (5)) Fx (1) =0.0276

Therefore, the real value tag of the sequence A1A1A3A2A3A1 is: T (113231) = (L (6) + U (6))/2=0.0273.

**6. For the probability model given in table 4-9, a serial number with a length of 10 labeled 0.63215699 is decoded.**

**Solution:**

Obtained from table: p (A1) =0.2, P (A2) =0.3, P (A3) =0.5

FX (0) =0,FX (1) =0.2, FX (2) =0.5, FX (3) =1.0, U (0) =1, L (0) =0

by X (AI) =i x (A1) =1,x (A2) =2,x (A3) =3

by formula, L (n) =l (n-1) + (U (n-1)-L (n-1)) Fx (xn-1)

U (n) =l (n-1) + (U (n-1)-L (n-1)) Fx (xn)

so: Set u (0) =1,L (0) =0

L (1) =0+ (1-0) Fx (x1-1) =fx (x1-1)

U (1) =0+ (1-0) Fx (x1) =fx (x1)

if x1=1, then the interval is [0,0.2]

if x1=2, then the interval is [0.2,0.5]

if x1=3, then the interval is [0.5,1]

1. because 0.63215699 is in [0.5,1] , x1=3

L (2) =0.5+ (1-0.5) Fx (x2-1) =0.5+0.5fx (x2-1)

U (2) =0.5+ (1-0.5) Fx (x2) =0.5+0.5fx (x2)

if x2=1, then the interval is [0.5,0.6]

if x2=2, then the interval is [0.6,0.75]

if x2=3, then the interval is [0.75,1]

2. because 0.63215699 is in [0.6,0.75] , x2=2

L (3) =0.5+ (0.75-0.6) Fx (x3-1) =0.6+0.15fx (x3-1)

U (3) =0.5+ (0.75-0.6) Fx (x3) =0.6+0.15fx (x3)

if x3=1, then the interval is [0.6,0.63]

if x3=2, then the interval is [0.63,0.675]

if x3=3, then the interval is [0.675,0.75]

3. because 0.63215699 is in [0.63,0.675] , x3=2

L (4) =0.63+ (0.675-0.625) Fx (x4-1) =0.63+0.045fx (x4-1)

U (4) =0.63+ (0.675-0.625) Fx (x4) =0.63+0.045fx (x4)

if x4=1, then the interval is [0.63,0.639)

if x4=2, then the interval is [0.639,0.6525]

if x4=3, then the interval is [0.6525,0.675]

4. because 0.63215699 is in [0.63,0.639) , so x4=1

L (5) =0.63+ (0.639-0.63) Fx (x5-1) =0.63+0.009fx (x5-1)

U (5) =0.63+ (0.639-0.63) Fx (X5) =0.63+0.009fx (X5)

if x5=1, then the interval is [0.63,0.6318]

if x5=2, then the interval is [0.6318,0.6345]

if x5=3, then the interval is [0.6345,0.639]

5. because 0.63215699 is in [0.6318,0.6345] , x5=2

L (6) =0.6318+ (0.6345-0.6318) Fx (x6-1) =0.6318+0.0027fx (x6-1)

U (6) =0.6318+ (0.6345-0.6318) Fx (X6) =0.6318+0.0027fx (X6)

if x6=1, then the interval is [0.6318,0.63234]

if x6=2, then the interval is [0.63234,0.63315]

if x6=3, then the interval is [0.63315,0.63345]

6. because 0.63215699 is in [0.6318,0.63234) , so x6=1

L (7) =0.6318+ (0.63234-0.6318) Fx (x7-1) =0.6318+0.00054fx (x7-1)

U (7) =0.6318+ (0.63234-0.6318) Fx (x7) =0.6318+0.00054fx (X7)

if x7=1, then the interval is [0.6318,0.631908]

if x7=2, then the interval is [0.631908,0.63207]

if x7=3, then the interval is [0.63207,0.63234]

7. because 0.63215699 is in [0.63207,0.63234] , x7=3

** **L (8) =0.63207+ (0.63234-0.63207) Fx (x8-1) =0.63207+0.00027fx (x8-1)

U (8) =0.63207+ (0.63234-0.63207) Fx (x8) =0.63207+0.00027fx (x8)

if x8=1, then the interval is [0.63207,0.632124]

if x8=2, then the interval is [0.632124,0.632205]

if x8=3, then the interval is [0.632205,0.63234]

8. because 0.63215699 is in [0.632124,0.632205] , x8=2

L (9) =0.632124+ (0.632205-0.632124) Fx (x9-1) =0.632124+0.000081fx (x9-1)

U (9) =0.632124+ (0.632205-0.632124) Fx (x9) =0.632124+0.000081fx (x9)

if x9=1, then the interval is [0.632124,0.6321402]

if x9=2, then the interval is [0.6321402,0.6321645]

if x9=3, then the interval is [0.6321645,0.632205]

9. because 0.63215699 is in [0.6321402,0.6321645] , x9=2

L (Ten) =0.6321402+ (0.6321645-0.6321402) Fx (x10-1) =0.6321402+0.00006243fx (x10-1)

U (Ten) =0.6321402+ (0.6321645-0.6321402) Fx (x10) =0.6321402+0.00006243fx (x10)

if x10=1, then the interval is [0.6321402,0.63214506]

if x10=2, then the interval is [0.63214506,0.63215235]

if x10=3, then the interval is [0.63215235.0.6321645]

because 0.63215699 is in [0.63215235,0.6321645] , x10=3

so the sequence is 3221213223 that is a3a2a2a1a2a1a3a2a2a3

Assignment for the fourth time