To find the probability that a randomly chosen boy and a randomly chosen girl are both seniors from the high school band, we first need to find the total number of boys and girls, as well as the number of seniors in each group.
From the table, we see:
- Boys:
- Seniors = 9
- Girls:
- Seniors = 7
Now we can calculate the total number of boys and girls in the band:
- Total boys = 10 (Freshmen) + 7 (Sophomores) + 10 (Juniors) + 9 (Seniors) = 36
- Total girls = 8 (Freshmen) + 11 (Sophomores) + 9 (Juniors) + 7 (Seniors) = 35
Now we can calculate the probability:
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The probability of choosing a senior boy: \[ P(\text{Boy is Senior}) = \frac{\text{Number of senior boys}}{\text{Total number of boys}} = \frac{9}{36} = \frac{1}{4} \]
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The probability of choosing a senior girl: \[ P(\text{Girl is Senior}) = \frac{\text{Number of senior girls}}{\text{Total number of girls}} = \frac{7}{35} = \frac{1}{5} \]
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The combined probability that both the boy and girl chosen are seniors is given by the product of their individual probabilities: \[ P(\text{Both are Seniors}) = P(\text{Boy is Senior}) \times P(\text{Girl is Senior}) = \frac{1}{4} \times \frac{1}{5} = \frac{1}{20} \]
However, looking at the answer options, we realize that \(\frac{1}{20}\) wasn't directly listed. Let's re-check what's represented in the options.
Since \(\frac{1}{20}\) doesn't match any of the choices, it appears none of them reflect that probability adequately. However, the closest fraction to \(\frac{1}{20}\) should be clarified, as it does not exist directly in the options given: 14, 1/4, 15, 1/5, 19, 1/9, 120.
Given the confusion and assuming we overview the counts correctly, I'd present that \(\frac{1}{20}\) derived from well said calculations aligns with investigations. However, please, double-check the interpretations of the welcome solutions if needed.
Here, I'm confirming; thus finalized the probability is \(\frac{1}{20}\), no exact fit to the answers perhaps.
This summary clarifies correctness might establish well as in true math plausibility.