Suppose that two balanced dice are tossed repeatedly and the sum of the two uppermost faces is determined on each toss. What is the probability that we obtain

a) a sum of 3 before we obtain a sum of 7?

b) a sum of 4 before we obtain a sum of 7?

Printable View

- February 22nd 2008, 02:33 PMmathlete2Probability of an event
Suppose that two balanced dice are tossed repeatedly and the sum of the two uppermost faces is determined on each toss. What is the probability that we obtain

a) a sum of 3 before we obtain a sum of 7?

b) a sum of 4 before we obtain a sum of 7? - February 22nd 2008, 03:49 PMPlato
- February 23rd 2008, 12:57 AMCharbel
a sum of before a sum of on the th roll

[Math]

x \in Z\cap {\left[ {2,12} \right]}

[/tex]

[Math]

y \in Z\cap {\left[ {2,12} \right]}

[/tex]

[Math]

n \in Z^+\cup \{0\}

[/tex]

lol? - February 23rd 2008, 04:02 AMPlato
Not that .

- February 23rd 2008, 08:45 AMmathlete2
Well, see the answer is actually 1/4(a) and 1/3(b). I just don't know how they went about getting it?

- February 23rd 2008, 09:36 AMPlato