Question

the spinner below is spun twice. if the spinner lands on a border, that spin does not count and spin again. it is equally likely that the spinner will land in each of the six sectors. for each question below, enter your response as a reduced fraction. find the probability of spinning red on the first spin and cyan on the second spin. find the probability of spinning blue on the first spin and red on the second spin. find the probability of not spinning blue on either spin. (not blue on the first spin and not blue on the second spin.)

Explanation:

Step1: Calculate probability of red - first spin

There are 3 red sectors out of 6 total sectors. So the probability of spinning red on the first spin, $P(R_1)=\frac{3}{6}=\frac{1}{2}$.

Step2: Calculate probability of cyan - second spin

There is 1 cyan sector out of 6 total sectors. So the probability of spinning cyan on the second spin, $P(C_2)=\frac{1}{6}$.

Step3: Calculate probability of red - first and cyan - second

Since the two spins are independent events, we multiply the probabilities. $P(R_1\cap C_2)=P(R_1)\times P(C_2)=\frac{1}{2}\times\frac{1}{6}=\frac{1}{12}$.

Step4: Calculate probability of blue - first spin

There are 2 blue sectors out of 6 total sectors. So the probability of spinning blue on the first spin, $P(B_1)=\frac{2}{6}=\frac{1}{3}$.

Step5: Calculate probability of red - second spin

There are 3 red sectors out of 6 total sectors. So the probability of spinning red on the second spin, $P(R_2)=\frac{3}{6}=\frac{1}{2}$.

Step6: Calculate probability of blue - first and red - second

Since the two spins are independent events, we multiply the probabilities. $P(B_1\cap R_2)=P(B_1)\times P(R_2)=\frac{1}{3}\times\frac{1}{2}=\frac{1}{6}$.

Step7: Calculate probability of not - blue on first spin

The number of non - blue sectors is $6 - 2=4$. So the probability of not spinning blue on the first spin, $P(\overline{B_1})=\frac{4}{6}=\frac{2}{3}$.

Step8: Calculate probability of not - blue on second spin

The number of non - blue sectors is $6 - 2=4$. So the probability of not spinning blue on the second spin, $P(\overline{B_2})=\frac{4}{6}=\frac{2}{3}$.

Step9: Calculate probability of not - blue on either spin

Since the two spins are independent events, we multiply the probabilities. $P(\overline{B_1}\cap\overline{B_2})=P(\overline{B_1})\times P(\overline{B_2})=\frac{2}{3}\times\frac{2}{3}=\frac{4}{9}$.

Answer:

1. $\frac{1}{12}$
2. $\frac{1}{6}$
3. $\frac{4}{9}$