math
posted by Anonymous .
A sixsided die (with numbers 1 through 6) and an eightsided die (with numbers 1 through 8) are rolled. What is the probability that there is exactly one 6 showing? Express your answer as a common fraction.

For all probabilities, multiply individual probabilities. For eitheror probabilities, add.
(1/6 * 7/8) + (5/6 * 1/8) = ? 
2/5

1/4

There are 2 cases: the 6 is rolled either on the 6 or 8 sided die. Therefore, we have ((1/6)(7/8))+((5/6)(1/8))=1/4.

We proceed using casework. If there is exactly one 6 showing, then the 6 must be on the sixsided die or the eightsided die (but not both).
Case 1: 6 on the 6sided die and not on the 8sided die
The probability of getting a 6 with the sixsided die is 1/6, and the probability of not rolling a 6 with the eightsided die is 7/8. We multiply these to build up to the full case: $\dfrac16 \cdot \dfrac78 = \dfrac7{48}$.
Case 2: 6 not on the 6sided die and on the 8sided die
The probability of not getting a 6 with the sixsided die is 5/6, and the probability of rolling a 6 with the eightsided die is 1/8. Together, these have a probability of $\dfrac56 \cdot \dfrac18 = \dfrac5{48}$.
Now we add the probabilities from the separate cases. The probability that there is exactly one 6 is
\[ \frac{7}{48} + \frac{5}{48} = \frac{12}{48} = \boxed{\frac{1}{4}}.\]