To determine the probability that David gets a 1 on both the die and the spinner, we need to know the probabilities of each event separately and then multiply these probabilities together.
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Rolling a die: A standard die has 6 sides (numbered 1 through 6). The probability of rolling a 1 is: \[ P(\text{rolling a 1}) = \frac{1}{6} \]
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Spinning the spinner: The actual design of the spinner isn't specified here, but typically, if we assume it's a fair spinner divided into equal sections, we would also need the total number of sections on the spinner. For example, if the spinner were divided into 6 equal sections (numbered from 1 to 6, similar to the die), the probability of landing on a 1 would be: \[ P(\text{landing on 1}) = \frac{1}{6} \]
Assuming the spinner has 6 equal sections, the combined probability of both events occurring would be: \[ P(\text{rolling a 1 and landing on 1}) = P(\text{rolling a 1}) \times P(\text{landing on 1}) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \]
However, none of the responses provided include \(\frac{1}{36}\).
If the spinner had a different number of sections, we would need to adjust our calculations accordingly. For instance, if the spinner had 12 equal sections, the probability of landing on a 1 would be: \[ P(\text{landing on 1}) = \frac{1}{12} \] And the combined probability would then be: \[ P(\text{rolling a 1 and landing on 1}) = \frac{1}{6} \times \frac{1}{12} = \frac{1}{72} \]
It appears based on typical scenarios; however, most assumptions lead toward either a 1/36 or a 1/12 probability, but to finalize the solution, proper spinner information is needed.
Now, let's go through the options provided:
- \( \frac{1}{5} \)
- \( \frac{5}{12} \)
- \( \frac{1}{12} \)
- \( \frac{1}{24} \)
Considering a spinner with possibly 12 sections would give us that final probability, we can suggest that " \(\frac{1}{12}\) " could be plausible if either event had a different divisor altogether.
In conclusion, without exact details on the spinner, the most logical choice among the provided options, based either on a fair die or a spinner is \(\frac{1}{12}\).