al applied probability

上面個位飛天魏國大將軍張遼2條都答得不對。尤其第一條

9(a)
P(there is no error)
=(0.9)^8=0.4305

(ii)
P(there is an even number of error)
=8C2(0.9)^6(0.1)^2+8C4(0.9)^4(0.1)^4+8C6(0.9)^2(0.1)^6+(0.1)^8
=0.1534

(iii)
P(there is an odd number of error)
=8C1(0.9)^7(0.1)^1+8C3(0.9)^5(0.1)^3+8C5(0.9)^3(0.1)^5+8C7(0.9)(0.1)^7
=0.4161

(b)(i)
The probability of accept the word
=(0.9)^9+(0.1534+0.4161)(0.1)
=0.44437

P(does not accept the word)
=1-0.44437
=0.55563

(ii)
P(correctly transmitted|reject the word)
=(0.9)^8(0.1)/0.55563
=0.07775

P(incorrectly transmitted|accept the word)
=(0.1534+0.4161)(0.1)/0.44437
=0.1282

10
(a)
Since the draw does not effect the result of the game.
So
P(A is the player who wins the first game)
=0.3/(0.3+0.2)
=0.6

You should notice that in this question the probability of draw is 0.5, if not, then the draw will effect the result of the game !! Also, you cannot calculate part c .

2008-04-24 17:52:54 補充：
P{accept and no error} + P{(accept and odd # error in first 8 bits}+ P{accept and even # error in the first 8 bits}
= (0.9)^9 + 0.4161x0.1 (WHY?) + 0.1534x0.9 (WHY?)
= 0.5671

原來依家 AL 咁深!
9(b)(i)
P{accept} = P{accept and ("no error" or "error")}
= P{accept and no error} + P{accept and error}
= P{accept and no error} + P{(accept and ("odd # error" or "even # error" in first 8 bits)}

2008-04-23 21:36:18 補充：
= P{accept and no error} + P{(accept and odd # error in first 8 bits}+ P{accept and even # error in the first 8 bits}
= (0.9)^9 + 0.4161x0.1 + 0.1534x0.9
= 0.5671

2008-04-23 21:43:54 補充：
10(a)
P{A first win a game}
= P{A first win a game and ("A win the FIRST GAME" or "B win the FIRST GAME" or "draw in the FIRST GAME")}

2008-04-23 21:46:13 補充：
= P{A first win a game and A win the FIRST GAME} +P{A first win a game and B win the FIRST GAME) + P{A first win a game and draw in the FIRST GAME}
= P{A first win a game |A win the FIRST GAME}P{A win the FIRST GAME} + 0 + P{A first win a game | draw in the FIRST GAME}P{draw in the FIRST GAME}

2008-04-23 21:49:39 補充：
= 1x P{A win the FIRST GAME} + P{A first wins a game}P{draw in the FIRST GAME}
( The last term is obtained by the independence of each game)
= 0.3 + P{A first wins a game}x0.5

i.e. P{A first wins a game} = 0.3 + P{A first wins a game}x0.5

2008-04-23 21:59:48 補充：
Solving, we have P{A first wins a game} = 0.6

2008-04-25 00:51:03 補充：
If there is an odd error in the first 8 bits, then the total number of "1" in the first 8-bits change parity.
i.e. odd to even or even to odd
Now, if the parity checking bit has no error (probability 0.9) , then the word is rejected.

2008-04-25 00:52:28 補充：
However, if the parity checking bit has an error (probability 0.1), then the PARITY of the total number "1" of all 9 bits remains unchanged, i.e. odd to odd or even to even, and hence the word will then be accepted.

2008-04-25 00:55:54 補充：
Similarly, if the number of error in the first 8 bits is EVEN and the parity checking bit has no error (probability 0.9) , the parity of total number of "1" for all 9 bits is unchanged and hence the word will then be accepted.

2008-04-25 00:58:55 補充：
我實在很難想像 AL 會問的咁既問題，哩條泊得住今年條PI-10。