Hard Prob Question (1 Viewer)

[taco man]

Can someone do this for me plz
A game is played in which 2 coloured dice are rolled once. The 6 faces of the black die are numbered 5,7,8,10,11,14. the 6 faces of the white die are numbered 3,6,9,12,13,15. The player wins if the number on the black die is bigger than the number on the white die:
-how many games must be played before you have a 90% chance of winning at least one game?

thx

 

[withoutaface]

total number of dice combos is 6²
For 14 there are 5 ways it'll be > than that white
For 11, 3 ways
For 10, 3 ways
For 8, 2 ways
For 7, 2 ways
For 5, 1 way


=16/36=4/9 chance of winning

Hence a 5/9 chance of not winning.

By binomial probability the answer comes from:

0.1>= nCn (5/9)^n (the chance of all games losing)

and from there you can use guess and check or logs.

 

[...]

erk

they don't do binomial probability in 2u

 

[withoutaface]

Bleh, well then forget the nCn and just find the number of times you have to go to make sure the chance of all losses is less than 10%, that kind of logic would be in the 2u course I think...

 

[acmilan]

Yeah well the 2 unit way would just be to mutiply 0.5 n times, so just get rid of the nCn which is just 1 anyways.

 

[taco man]

I still dun get it , can u explain wat the question is asking for

 

[Riviet]

total number of dice combos is 6²
For 14 there are 5 ways it'll be > than that white
For 11, 3 ways
For 10, 3 ways
For 8, 2 ways
For 7, 2 ways
For 5, 1 way


=16/36=4/9 chance of winning

Hence a 5/9 chance of not winning.

I will complete the question without this binomial stuff, but you will need logs to do it this way:
Since there is 5/9 chance of not winning,
.: if i play 1 game, there is 5/9 chance of not winning
.: if i play 2 games, there is (5/9)^2 chance of not winning
.: if i play n games, there is (5/9)^n chance of not winning
we want to play a certain number of games to ensure that we have a 1/10 chance or less of losing
.: (5/9)^n < 1/10
n log (5/9) < log (1/10)
n < log(1/10) / log (5/9)
.: n < 3.917.....
This means if i played 4 games, there would be about a 9.5% chance of losing or 90.5% of winning which is what we were required to find
So you would need to play the game 4 times to ensure a 90% success rate.

P.S I think this is right, correct or criticise me if im wrong

 

[taco man]

nah its rite, says 4 in the answers
thanks

 

[Riviet]

Woohoo!! I hope that will help you get a slightly higher mark in the hsc exam. Good luck mate!