# Dice -_-;

• Jul 21st 2010, 12:23 AM
Dave2718
Dice -_-;
Here's the question:

In rolling a fair die, Find the probability of rolling a 1 before an even number? All I'm given as a hint is to consider

P(1|1 or even)

I know it's conditional probability, I just don't know if i need to use Bayes' theorem or something different, i know the answer is 1/4 but for some reason I keep coming up with 1/2, which cannot be right at all. I feel like I'm making a really dumb mistake and have been working on this one for hours.
• Jul 21st 2010, 01:10 AM
SpringFan25
you are going to keep rolling the die until you get a 1 or an even number (since at that point, you know the answer).

The chance of it being a 1 on that roll is P(1|1 or even)

which is 1/4

(since there are 4 values of the die that are 1 or even : 1,2,4,6)
• Jul 21st 2010, 11:07 AM
Soroban
Hello, Dave2718!

That hint was a good one . . .
. . and SpringFan25 is absolutely correct.

Quote:

In rolling a fair die, find the probability of rolling a 1 before an even number?

We can also solve this "the long way" . . .

We are concerned with 3 events:

. . $\begin{array}{ccccc}P(\text{One}) &=& \frac{1}{6} \\ \\[-3mm]
P(\text{Even}) &=& \frac{3}{6} \\ \\[-3mm]
P(\text{Other}) &=& \frac{2}{6} \end{array}$

Then we have these possible outcomes:

. . $\begin{array}{cccccccc}
P(\text{One on 1st roll}) &=& \frac{1}{6} \\ \\[-4mm]
P(\text{One on 2nd roll}) &=& \left(\frac{2}{6}\right)\cdot\frac{1}{6} \\ \\[-4mm]
P(\text{One on 3rd roll}) &=& \left(\frac{2}{6}\right)^2\cdot \frac{1}{6} \\ \\[-4mm]
P(\text{One on 4th roll}) &=& \left(\frac{2}{6}\right)^3\cdot\frac{1}{6} \\ \\[-4mm]
\vdots && \vdots \end{array}$

$P(\text{One before Even}) \;=\;\frac{1}{6} + \frac{1}{6}\left(\frac{1}{3}\right) + \frac{1}{6}\left(\frac{1}{3}\right)^2 + \frac{1}{6}\left(\frac{1}{3}\right)^3 + \hdots$

. . . . . . . . . . . . . . . . $=\;\frac{1}{6}\underbrace{\bigg[1 + \tfrac{1}{3} + \left(\tfrac{1}{3}\right)^2 + \left(\tfrac{1}{3}\right)^3 + \hdots \bigg]}_{\text{geometric series}}$

The sum of the geometric series is: . $\dfrac{1}{1-\frac{1}{3}} \:=\:\dfrac{1}{\frac{2}{3}} \:=\:\frac{3}{2}$

Therefore: . $P(\text{One before Even}) \;=\; \frac{1}{6}\cdot\frac{3}{2} \;=\;\boxed{\frac{1}{4}}$