About Discrete Probability~ Help!

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Main Question or Discussion Point

Zaki and Ramli play a game in which they take it in turns to toss a tetrahedral dice. They agree that the first man to toss a "2" wins the game. Ramli toesses the dice first.

(a) Find the probability that Zaki loses on his first toss. (Ans : 3/16)

(b) If x is the probability that Ramli wins the game, show that x = 1/4 + 9/16 x

(c) Ramli and Zaki plays the game five times. Find the probability that Zaki wins more games than Ramli. (Ans :0.368)
 

Answers and Replies

  • #2
tiny-tim
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Hi Scharles! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
  • #3
mathman
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What are the numbers on the tetrahedral die?
The (a) question is confusing. The game seems to be such that the tosser may win or continue, so Zaki could lose only when Ramli tosses, not when he (Zaki) tosses.
 
  • #4
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for (a), i used 2 method to solve this question, first method get the answer but second method the answer i get was different ~ can u tell me which part of my solution was wrong ?

method 1 :
let R = ramli win and Z = zaki win

P(Zaki loses on his first toss) = P(R'Z'R) + P(R'Z'R'Z) + P(R'Z'R'Z'R) + ......
= (3/4)(3/4)(1/4) + (3/4)(3/4)(3/4)(1/4) + .......
= (9/64) (1 + 1/4 + 1/4*2 + ........)
=(9/64) (1/[1-1/4]) (sum of infinity)
= (9/64)(4/3)
= 3/16

second method,
P(Zaki loses on his first toss)= 1 - P(R) - P(R'Z)
= 1 - 1/4 - (3/4)(1/4)
= 9/16
 
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  • #5
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for (b), i do like this .......

P(x) = P(R) + P(R'Z'R) + P(R'Z'R'Z'R) + .....
= 1/4 + (3/4)(3/4)(1/4) + (3/4)(3/4)(3/4)(3/4)(1/4) + ....
= (1/4) ( 1 + [9/16] + [9/16]*2 + .......)
= (1/4) (1/[1-9/16]) sum of infinity..
= (1/4)(16/7)
= 4/7

the answer i get is the whole number which does not contain any unknown x, may i know how to show it by using the correct method ?
 
  • #6
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for (c), i have no idea, but i had tried it using binomial distribution...

Let Z = Zaki wins the game

Z~ B ( 5, 0.25)
P(Z>2) = P(Z=3) + P(Z=4) + P(Z=5)
= 5C3(0.25*3)(0.75*2) + 5C4(0.25*4)(0.75) + (0.25*5)
= 0.1035

which is not same to the answe given~ am i using a wrong method to do ??
 
  • #7
tiny-tim
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Hi Scharles! :smile:

(try using the X2 icon just above the Reply box :wink:)
Zaki and Ramli play a game in which they take it in turns to toss a tetrahedral dice. They agree that the first man to toss a "2" wins the game. Ramli toesses the dice first.

(a) Find the probability that Zaki loses on his first toss. (Ans : 3/16)
I don't understand … you can only win on a toss … how cam Zaki lose on a toss?

Please explain the game more clearly. :confused:
method 1 :
let R = ramli win and Z = zaki win

P(Zaki loses on his first toss) = P(R'Z'R) + P(R'Z'R'Z) + P(R'Z'R'Z'R) + ......
= (3/4)(3/4)(1/4) + (3/4)(3/4)(3/4)(1/4) + .......
= (9/64) (1 + 1/4 + 1/4*2 + ........)
=(9/64) (1/[1-1/4]) (sum of infinity)
= (9/64)(4/3)
= 3/16
As I said above, I dont understand (a), but anyway …
i] your series should be (1 + 3/4 + 3/42 + ........)
ii] your first line is the same as 1 - P(R) - P(R'Z)

second method,
P(Zaki loses on his first toss)= 1 - P(R) - P(R'Z)
= 1 - 1/4 - (3/4)(1/4)
= 9/16
No, 1 - P(R) - P(R'Z) = P(R'Z'). :redface:
 

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