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.
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 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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1/4
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To find the probability of getting exactly one 6, we need to consider the possible outcomes where exactly one 6 is showing, and then divide that by the total number of possible outcomes.
Let's start by finding the total number of possible outcomes when rolling the two dice. The six-sided die has 6 possible outcomes, and the eight-sided die has 8 possible outcomes. Since the two dice are independent, the total number of outcomes when rolling both dice is the product of the number of outcomes for each die:
Total number of outcomes = 6 outcomes (from the six-sided die) * 8 outcomes (from the eight-sided die)
= 48 outcomes
Now let's consider the possible outcomes where exactly one 6 is showing.
We have two cases to consider:
1. The six-sided die shows a 6, and the eight-sided die does not.
2. The eight-sided die shows a 6, and the six-sided die does not.
For case 1, the probability of the six-sided die showing a 6 is 1/6, and the probability of the eight-sided die not showing a 6 is 7/8 (since there are 7 other numbers on the eight-sided die). So, the probability for case 1 is:
Probability of case 1 = (1/6) * (7/8) = 7/48
Similarly, for case 2, the probability of the eight-sided die showing a 6 is 1/8, and the probability of the six-sided die not showing a 6 is 5/6. So, the probability for case 2 is:
Probability of case 2 = (1/8) * (5/6) = 5/48
To find the total probability of exactly one 6 showing, we sum up the probabilities of the two cases:
Total probability of exactly one 6 showing = Probability of case 1 + Probability of case 2
= 7/48 + 5/48
= 12/48
= 1/4
Therefore, the probability that there is exactly one 6 showing when rolling a six-sided die and an eight-sided die is 1/4.
For all probabilities, multiply individual probabilities. For either-or probabilities, add.
(1/6 * 7/8) + (5/6 * 1/8) = ?