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 cyan on the second spin. find the probability of not spinning red on either spin. (not red on the first spin and not red on the second spin.)

Explanation:

Step1: Calculate probability of first - spin event

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

Step2: Calculate probability of second - spin event

The probability of spinning cyan on the second spin, $P(C_2)=\frac{1}{6}$. Since the spins are independent events, the probability of spinning red on the first spin and cyan on the second spin is $P(R_1\cap C_2)=P(R_1)\times P(C_2)=\frac{1}{2}\times\frac{1}{6}=\frac{1}{12}$.

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

There are 2 blue sectors. The probability of spinning blue on the first spin, $P(B_1)=\frac{2}{6}=\frac{1}{3}$. The probability of spinning cyan on the second spin, $P(C_2)=\frac{1}{6}$. Since the spins are independent, $P(B_1\cap C_2)=P(B_1)\times P(C_2)=\frac{1}{3}\times\frac{1}{6}=\frac{1}{18}$.

Step4: Calculate probability of not - red on each spin

The probability of not spinning red on a single spin is $P(\text{not }R)=\frac{6 - 3}{6}=\frac{3}{6}=\frac{1}{2}$. Since the spins are independent, the probability of not spinning red on either spin is $P(\text{not }R_1\cap\text{not }R_2)=P(\text{not }R_1)\times P(\text{not }R_2)=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$.

Answer:

The probability of spinning red on the first spin and cyan on the second spin is $\frac{1}{12}$.
The probability of spinning blue on the first spin and cyan on the second spin is $\frac{1}{18}$.
The probability of not spinning red on either spin is $\frac{1}{4}$.