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

Explanation:

Step1: Calculate probability of blue on first - spin and red on second - spin

There are 2 blue sectors out of 6, so the probability of spinning blue on the first spin, $P(B_1)=\frac{2}{6}=\frac{1}{3}$. There are 3 red sectors out of 6, so the probability of spinning red on the second spin, $P(R_2)=\frac{3}{6}=\frac{1}{2}$. Since the two spins are independent events, the probability of both events occurring is the product of their probabilities. So $P(B_1\cap R_2)=P(B_1)\times P(R_2)=\frac{1}{3}\times\frac{1}{2}=\frac{1}{6}$.

Step2: Calculate probability of cyan on first - spin and cyan on second - spin

There is 1 cyan sector out of 6, so the probability of spinning cyan on the first spin, $P(C_1)=\frac{1}{6}$, and the probability of spinning cyan on the second spin, $P(C_2)=\frac{1}{6}$. Since the spins are independent, $P(C_1\cap C_2)=P(C_1)\times P(C_2)=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$.

Step3: Calculate probability of not - cyan on either spin

The probability of not spinning cyan on a single spin is $P(\text{not }C)=1 - P(C)$. Since $P(C)=\frac{1}{6}$, then $P(\text{not }C)=1-\frac{1}{6}=\frac{5}{6}$. Since the two spins are independent, the probability of not spinning cyan on either spin is $P(\text{not }C_1\cap\text{not }C_2)=P(\text{not }C_1)\times P(\text{not }C_2)=\frac{5}{6}\times\frac{5}{6}=\frac{25}{36}$.

Answer:

$\frac{1}{6}$
$\frac{1}{36}$
$\frac{25}{36}$