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?

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- Feb 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? - Feb 22nd 2008, 03:49 PMPlato
Consider the probabilities $\displaystyle P(3) = \frac{2}{{36}}\,\& \,P(7) = \frac{6}{{36}}$.

In order for there to be a sum of three before a sum of seven happening on say the fifth toss we must have**four sums of**. That probability is $\displaystyle \left( {\frac{{28}}{{36}}} \right)^4 \left( {\frac{2}{{36}}} \right)$.__neither__three nor seven and than a sum of three on the fifth toss

Will this work $\displaystyle \sum\limits_{k = 1}^\infty {\left( {\frac{{28}}{{36}}} \right)^{k - 1} \left( {\frac{2}{{36}}} \right)} $? - Feb 23rd 2008, 12:57 AMCharbel
a sum of $\displaystyle x$ before a sum of $\displaystyle y$ on the $\displaystyle n$th roll

$\displaystyle

\left[ {1 - \left( {\frac{{ - \left| {y - 7} \right| - \left| {x - 7} \right| + 12}}

{{36}}} \right)} \right]^{n - 1} \left[ {\frac{{ - \left| {x - 7} \right| + 6}}

{{36}}} \right]

$

[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? - Feb 23rd 2008, 04:02 AMPlato
Not that $\displaystyle n \in Z^ + \, \mbox{not} \,n \in \left[ {Z^ + \cup \left\{ 0 \right\}} \right]$.

- Feb 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?

- Feb 23rd 2008, 09:36 AMPlato