Question

Bruno will spin the arrow on the spinner 2 times.

What is the probability that the arrow will stop on C at least once?

Responses

19
1 over 9

29
2 over 9

13
1 over 3

59

Answer 1

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.

1. **Calculate the probability of not landing on C in a single spin:**
\[
P(\text{not C}) = 1 - P(C) = 1 - \frac{1}{n}
\]

2. **Calculate the probability of not landing on C in two spins:**
The probability of not landing on C in two consecutive spins is:
\[
P(\text{not C in 2 spins}) = P(\text{not C}) \times P(\text{not C}) = \left(1 - \frac{1}{n}\right)^2
\]

3. **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.