To find the probability that the arrow will stop on segment C at least once when spun twice, we can use the complementary probability approach.
First, we need to define the probability of the arrow landing on C in a single spin. Let's assume the spinner has \( n \) equal sections and one of those sections is C.
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Calculate the probability of not landing on C in a single spin: \[ P(\text{not C}) = 1 - P(C) = 1 - \frac{1}{n} \]
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Calculate the probability of not landing on C in two spins: The probability of not landing on C in two consecutive spins is: \[ P(\text{not C in 2 spins}) = P(\text{not C}) \times P(\text{not C}) = \left(1 - \frac{1}{n}\right)^2 \]
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Calculate the probability of landing on C at least once in two spins: The probability of landing on C at least once is then: \[ P(\text{at least one C}) = 1 - P(\text{not C in 2 spins}) = 1 - \left(1 - \frac{1}{n}\right)^2 \]
Now, without knowing how many sections \( n \) the spinner has, we cannot numerically calculate the probabilities. However, we can analyze the given options:
- 19
- \( \frac{1}{9} \)
- 29
- \( \frac{2}{9} \)
- \( \frac{1}{3} \)
- 59
Assuming the spinner has 3 equal sections (A, B, C), then \( n = 3 \).
Calculating the probabilities:
- \( P(C) = \frac{1}{3} \)
- \( P(\text{not C}) = 1 - \frac{1}{3} = \frac{2}{3} \)
Then, \[ P(\text{not C in 2 spins}) = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \]
Thus, \[ P(\text{at least one C}) = 1 - \frac{4}{9} = \frac{5}{9} \]
This alignment with content choices is not visible, but if sections reduce to fractions, summation leads us to the approximated \( 2/9 \) section selected previously.
Given the selection process in responses:
- The closest reasonable option given practical sections and practice would be \( \frac{1}{3} \) (assumed rounded).
In summation, check for how many sections are included statistically. If actual sections differ, adjust \( n \) and recalculate respective probabilities directly relating to selections best matching probable outcomes.