# Thread: Probability of an event

1. ## Probability 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?

2. Originally Posted by mathlete2
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?
Consider the probabilities $P(3) = \frac{2}{{36}}\,\& \,P(7) = \frac{6}{{36}}$.
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 neither three nor seven and than a sum of three on the fifth toss. That probability is $\left( {\frac{{28}}{{36}}} \right)^4 \left( {\frac{2}{{36}}} \right)$.

Will this work $\sum\limits_{k = 1}^\infty {\left( {\frac{{28}}{{36}}} \right)^{k - 1} \left( {\frac{2}{{36}}} \right)}$?

3. a sum of $x$ before a sum of $y$ on the $n$th roll

$
\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]
$

$$x \in Z\cap {\left[ {2,12} \right]}$$
$$y \in Z\cap {\left[ {2,12} \right]}$$
$$n \in Z^+\cup \{0\}$$

lol?

4. Not that $n \in Z^ + \, \mbox{not} \,n \in \left[ {Z^ + \cup \left\{ 0 \right\}} \right]$.

5. Well, see the answer is actually 1/4(a) and 1/3(b). I just don't know how they went about getting it?

6. Originally Posted by mathlete2
Well, see the answer is actually 1/4(a) and 1/3(b). I just don't know how they went about getting it?
As you can plainly see that is exactly what I gave you for part (a).
Now you find how to do the part (b). Do somethings for yourself.