Question

write all answers as fractions (do not simplify).
suppose you spin the spinner one time.
a) what is the probability that the spinner lands on red?
p(red) =

b) what is the probability that the spinner does not land on red?
p(not red) =

c) what is the probability that the spinner lands on green or yellow?
p(green or yellow) =

suppose you spin the spinner two times.
d) what is the probability that the spinner lands on red both times?
p(red and red) =

e) what is the probability that the spinner lands on blue first and then yellow?
p(blue and yellow) =

Explanation:

Step1: Count total sections

The spinner has 12 sections.

Step2: Count red sections for part a

There are 6 red sections. So \(P(\text{red})=\frac{6}{12}\)

Step3: Use complement rule for part b

\(P(\text{not red}) = 1 - P(\text{red})\). Since \(P(\text{red})=\frac{6}{12}\), then \(P(\text{not red})=\frac{12 - 6}{12}=\frac{6}{12}\)

Step4: Count green and yellow sections for part c

There are 3 green and 1 yellow section. Total of \(3 + 1=4\) sections. So \(P(\text{green or yellow})=\frac{4}{12}\)

Step5: Use multiplication rule for independent events (part d)

Since spins are independent, \(P(\text{red and red})=P(\text{red})\times P(\text{red})\). \(P(\text{red})=\frac{6}{12}\), so \(P(\text{red and red})=\frac{6}{12}\times\frac{6}{12}\)

Step6: Use multiplication rule for independent events (part e)

There are 2 blue sections (\(P(\text{blue})=\frac{2}{12}\)) and 1 yellow section (\(P(\text{yellow})=\frac{1}{12}\)). \(P(\text{blue and yellow})=P(\text{blue})\times P(\text{yellow})=\frac{2}{12}\times\frac{1}{12}\)

Answer:

a) \(\frac{6}{12}\)
b) \(\frac{6}{12}\)
c) \(\frac{4}{12}\)
d) \(\frac{6}{12}\times\frac{6}{12}\)
e) \(\frac{2}{12}\times\frac{1}{12}\)