Veronica rolls 2 dice. As long as she rolls the same numbers twice, she solls it again.What is the chance that she rolls 9 as the last throw?
9 is the sum of
4,5
6,3
There are 4 ways this can occur when 2 dice are rolled.
The total number of ways two dice can be rolled is 36
So 4/36 chance of rolling a sum of 9 : )
And reduce that and change it to a percent : )
Move on to the next question in your homework... I will work on your current question and see if we can come up with that answer...
Note... our answer reduces to 2/18 : )
Which means there could be a typo in the answers...
I will give it more thought : )
Great perseverance for getting your work done!!
The "game" ends when you roll a 9
As Ms Pi noted (Happy belated π Day), the prob(9) = 4/36 = 1/9
The game could with with
- a roll of 9 to start with
- a roll of doubles, then a 9
- a roll of 2 doubles, then a 9
- a roll of 3 doubles, then a 9
.... for "quite a few more"
P(end game)
= (1/9) + (1/9)(1/6) + (1/9)(1/6)^2 + (1/9)(1/6)^3 + ....
= (1/9)[ 1 + 1/6 + (1/6)^2 + (1/6)^3 + ..... ]
the part in the bracket is an infinite GS , where a = 1 and r = 1/6
sum = a/(1-r) = 1/(5/6) = 6/5
= (1/9)(6/5) = 2/15
Véronique rolls 2 dice. As long as she rolls double (same eyes twice) she rolls again.
1)What is the chance that she rolls 9 as the last throw?