Help with 2U Probability !!!!!! (1 Viewer)

[flowerp]

Hi everyone,
I need help with the following question please:

David has invented a game for one person. He throws two ordinary dice repeatedly until the sum of the two numbers shown is either 7 or 9. If the sum is 9, David wins. If the sum is 7, David loses. If the sum is any other number, he continues to throw until it is 7 or 9.

(i) Show that the probability that David wins on his first throw of the dice is 1/9.

(ii) Calculate the probability that a second throw is needed.

(iii) What is the probability that David wins on his first, second, or third throw? Leave your answer in unsimplified form.

(iv) Calculate the probability that David wins the game.

THANK YOU HEAPS!!!!

 

[StudyOnly]

10(a)
i)
...1 2 3 4 5 6
1 ...............x
2 ............x...
3 .........x.....o
4 ......x......o..
5....x.....o......
6 x....o..........

Chance of winning = 4/36 = 1/9

ii)
Second throw is needed when you get neither the "win" or "lose" = 26/36 = 13/18

iii)

David winning first throw = 1/9
David winning second throw = 13/18 . 1/9 (because he shouldn't have either won or lost in the first throw)
David winning thrid throw = 13/18 . 13/18 . 1/9 = 169/2916

Adding them altogether: 727/2916

iv) The series is formed as you see my above working out. It's an infinite sum since r = 13/18 < 1

S = a/(1-r) = 1/9/(1-13/18) = 2/5

 

[flowerp]

THANK YOU SO MUCH !!!!

 

[StudyOnly]

THANK YOU SO MUCH !!!!

You should thank Iyounamu lol, but I guess you're welcome for finding it.
If you ever have a question try googling the question first, more often than not someone has already asked it before.

 

[flowerp]

will do in the future