1. ## dice

suppose that 2 balanced dice are tossed repeatedly and the sum of the 2 uppermost faces is determined on each toss. what is the porbabilty that we obtain a sum of 3 before we obtain a sum of 7.

my working:

since obtain 3 and 7 are independent,
P( getting sum of 3) = 2/6 x 1/6
P ( getings sum of 7) = 3/6 x 1/6

so the answer should be 2/6 x 1/6 x 2/6 x 1/6

but my book says taht it is 1/4...

2. Originally Posted by stressedout
suppose that 2 balanced dice are tossed repeatedly and the sum of the 2 uppermost faces is determined on each toss. what is the porbabilty that we obtain a sum of 3 before we obtain a sum of 7.

my working:

since obtain 3 and 7 are independent,
P( getting sum of 3) = 2/6 x 1/6
P ( getings sum of 7) = 3/6 x 1/6

so the answer should be 2/6 x 1/6 x 2/6 x 1/6

but my book says taht it is 1/4...
This one's actually a little tricky. What you seem to be calculating is the probability that you'll get a sum of 3 on the first throw, and a sum of 7 on the second throw (or vice versa). What you actually want is the probability that you attain a sum of 3 before a sum of 7.

To begin with, determine the respective probabilities:

$\displaystyle P$(sum of 3 on any roll of the dice)$\displaystyle = \frac{2}{36}$

$\displaystyle P$(sum of 7 on any roll of the dice) $\displaystyle = \frac{6}{36}$

(You should be able to obtain these easily by counting the outcomes).

Now, we want the probability of attaining sum of 3 before sum of 7. This is equivalent to the probability that, when you obtain the first sum of 3, your previous rolls have not had sums of 7 (or 3, obviously). In this sense we have what resembles a geometric random variable:

$\displaystyle P$(first sum of 3 obtained on the $\displaystyle x^{th}$ roll, without obtaining any sums of 7 before the $\displaystyle x^{th}$ roll)$\displaystyle = (\frac{28}{36})^{x-1}\frac{2}{36}$.

This is of course for only some value of $\displaystyle x$. Summing over all possible values of $\displaystyle x$, we get:

$\displaystyle P$(first sum of 3 obtained without obtaining sums of 7 before) $\displaystyle = \sum_{x=1}^\infty (\frac{28}{36})^{x-1}\frac{2}{36} = \frac{\frac{2}{36}}{1 - \frac{28}{36}} = \frac{1}{4}$

3. supposed it is to get a sum of 4 before a sum of 7, the method would be the same just that instead of 2/36 in the final equation, it will be 3/ 36 divide by 8/ 36 right?

4. Originally Posted by stressedout
supposed it is to get a sum of 4 before a sum of 7, the method would be the same just that instead of 2/36 in the final equation, it will be 3/ 36 divide by 8/ 36 right?
You've got the right idea with $\displaystyle \frac{3}{36}$, but remember you have to adjust the other denominator accordingly.

Probability of not rolling a sum of 7 or 4 is $\displaystyle 1 - \frac{3}{36} - \frac{6}{36} = \frac{27}{36}$

Hence you want to divide $\displaystyle \frac{\frac{3}{36}}{\frac{9}{36}}$.