Question

All odd numbers are equally likely, all the even numbers are equally likely, and the odd numbers are k times as likely as the even numbers, and Pr[1] = [ 2/7]. What is k?

Answer 1

What is meant by \(\Pr[1]=[\frac{2}{7}]\)?

Answer 2

The probability of rolling a one (an odd number) is 2/7. So the other odd numbers also have a probability of 2/7 since they are all equa

Answer 3

*equal

Answer 4

All right, then we take that:
\[
\frac{2}{7}=(1-\frac{2}{7})k
\]And solve for \(k\), since the probabilities must add up to 1.

Answer 5

that would mean k would be 2/5, but I found the answer to be 6, I just don't remember how I found it.

Answer 6

The answer can't be 6, because \(6^{-1}\cdot \frac{2}{7}+\frac{2}{7}\neq1\)

Answer 7

Ahh! I'm confused. I know the answer is 6, but I don't remember how to get it!

Answer 8

The answer cannot be 6, unless there is an error in the problem above.

Answer 9

It says the answer is 6!!

Answer 10

Maybe I'm just stupid... wouldn't surprise me. But I cannot see how it could be. Since \(6\frac{2}{7}+\frac{2}{7}>1\) as is \(6\frac{5}{7}+\frac{5}{7}\), and, by the third choice, the case above.