Question

A spinner has 8 equal-sized sections. Four of the sections are blue,
two are red, and two are green. The pointer lands in a red section
8 times in 20 spins. How does this compare to the number of times
the pointer is expected to land in a red section?
Step 1 Use the experiment’s results to find the experimental
probability that the pointer lands in a red section.
Experimental probability = number of times pointer lands in a red section
total number of spins = 8
20 = 40%
Step 2 Find the theoretical probability that the pointer lands in a
red section.
P(red) = number of red sections
total number of equal@sized sections = 2
8 = 25%
The experimental probability is greater than the theoretical probability.
The pointer landed in a red section more often than expected.
A spinner has 10 equal-sized sections labeled 1 through 10. In 40 spins,
the spinner lands 3 times in section 5. How does this compare to the
number of times the pointer is expected to land in section 5?
1. Find the experimental probability that the pointer lands in section 5.
number of times pointer lands in section 5
total number of spins = □ = %
2. Find the theoretical probability that the pointer lands in section 5.
P(5) = number of sections labeled "5"
total number of equal@sized sections = □ = %
3. How does the actual number of times the pointer landed in section 5
compare to the expected number?
On the Back!
4. A spinner has 4 equal-sized sections labeled 1 through 4. In 25 spins,
the spinner lands 5 times in section 3. How does this compare to the
number of times the pointer is expected to land in section 3?

Answer 1

Second spinner (10 sections, 40 spins, 3 lands on 5)

1. Experimental probability = 3/40 = 0.075 = 7.5%.

2. Theoretical probability P(5) = 1/10 = 10%.

3. Expected number in 40 spins = 40 × (1/10) = 4. Actual = 3, which is 1 less than expected. So the pointer landed in section 5 fewer times than expected.

Fourth question (4 sections, 25 spins, 5 lands on 3)

- Experimental probability = 5/25 = 1/5 = 20%.
- Theoretical probability P(3) = 1/4 = 25%.
- Expected number in 25 spins = 25 × (1/4) = 6.25. Actual = 5, which is 1.25 fewer than expected, so it landed in section 3 less often than expected.