# math

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A six-sided die (with numbers 1 through 6) and an eight-sided 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.

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For all probabilities, multiply individual probabilities. For either-or probabilities, add.

(1/6 * 7/8) + (5/6 * 1/8) = ?

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2/5

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1/4

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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.

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We proceed using casework. If there is exactly one 6 showing, then the 6 must be on the six-sided die or the eight-sided die (but not both).

Case 1: 6 on the 6-sided die and not on the 8-sided die
The probability of getting a 6 with the six-sided die is 1/6, and the probability of not rolling a 6 with the eight-sided 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 6-sided die and on the 8-sided die
The probability of not getting a 6 with the six-sided die is 5/6, and the probability of rolling a 6 with the eight-sided 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
$\frac{7}{48} + \frac{5}{48} = \frac{12}{48} = \boxed{\frac{1}{4}}.$

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