A 6-sided die is weighted so that the probabilities of rolling a 1, 2, or 3 are equally likely; the probabilities of rolling a 4, 5, or 6 are equally likely; and the probability of rolling a 5 is four times the probability of rolling a 3. What is the probability that a 2 or a 5 is rolled?
How I solved it:
$P(\text{rolling a } 1, 2, 3) = x$
$P(\text{rolling a } 4, 5, 6) = 4x$ (because rolling a 4, 5, 6 is 4x the probability of rolling a 1, 2, 3)
$3(x)+3(4x)=15x$ (because 1,2,3 each has $x$ probability of being rolled and same with 4,5,6 except each of them has a $4x$ chance)
$15x = 1 , x=\frac{1}{15}$
Therefore, there's a $\frac{1}{15}$ chance of rolling a 1, rolling a 2, rolling a 3 and a $\frac{4}{15}$ chance of rolling a 4, rolling a 5, rolling a 6.
So to roll a 2 or 5, that would be $\frac{1}{15} + \frac{4}{15} = \boxed{\frac{1}{3}}$.
My teacher told me I got the answer correct, but my solution wasn't right and she said the part where it mentions that rolling a 5 is four times the probability of rolling a 3 is irrelevant. Can someone tell me what I'm doing wrong? Thanks!
Best Answer
Your solution was correct but there was an easier way to see it.
The probability of a $1$ or $4$ is the same as the probability of a $2$ or $5$, which is the same as the probability of a $3$ or $6$. Thus, the answer has to be $\frac 13$ no matter what the relative probability of the two groups are.