To find the probability of landing on an unshaded section of a spinner labeled with the letters of the word "HONEST", we first need to determine the total number of sections and which of those sections are shaded and unshaded.
The word "HONEST" has 6 letters: H, O, N, E, S, T. Assuming that all sections (letters) on the spinner are equally likely and that there are no additional specifications about shading, we can proceed as follows:
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Total Sections: There are 6 sections (one for each letter).
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Assumption for Shading: Since the problem does not specify which letters are unshaded or how many, we'll assume that all letters are unshaded unless otherwise indicated.
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Determining Probability: If all sections are unshaded:
- The probability of landing on an unshaded section is the number of unshaded sections divided by the total number of sections.
- All sections are unshaded, so the probability is: \[ \text{Probability} = \frac{\text{Number of unshaded sections}}{\text{Total sections}} = \frac{6}{6} = 1 \]
However, if we consider the possibility of some sections being shaded, we need clear information on how many are unshaded. Without loss of generality, if we analyze the provided options assuming some sections might be shaded:
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If we denote with "U" the number of unshaded sections and "S" the number of shaded sections, we have \( U + S = 6 \). Without specific numbers for U or S, the options provided can be examined for validity.
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If 1 section is shaded (5 unshaded), the probability becomes \(\frac{5}{6}\).
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If 2 sections are shaded (4 unshaded), the probability is \(\frac{4}{6} = \frac{2}{3}\).
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If 3 sections are shaded (3 unshaded), the probability is \(\frac{3}{6} = \frac{1}{2}\).
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If 4 sections are shaded (2 unshaded), the probability is \(\frac{2}{6} = \frac{1}{3}\).
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If 5 sections are shaded (1 unshaded), the probability is \(\frac{1}{6}\).
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If all 6 are shaded (0 unshaded), the probability = 0.
Matching with available options:
Given these derivations, let's match against the options:
- \(1\) (not valid unless all are unshaded)
- \(\frac{1}{6}\) (valid if 5 shaded)
- \(\frac{1}{2}\) (valid if 3 shaded)
- \(\frac{1}{3}\) (valid if 4 shaded)
- \(3\) (not feasible as a probability)
Since there's no provided specific number of unshaded versus shaded, let's choose the option that makes sense bulks of positive probabilities based on general practices:
Final Answer (Assuming uniform distribution without specific indication):
- If interpreting that the spinner displays all letters unshaded, the probability is \(1\).
- As it relates to options, the most reasonable given usual shading probabilities would lead us to select from the residual \(\frac{1}{2}\) or \(\frac{1}{3}\) based on assumptions of common shading practices.
Select: \(\frac{1}{2}\) or \(\frac{1}{3}\) based upon interpretation of shaded sections (assuming even distribution and limited shading specified).