Questions: 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. Find the probability of NOT spinning red on either spin. (Not red on the first spin and not red on the second spin.)

Solution Steps

The spinner has 6 sectors: 3 red, 2 blue, and 1 cyan. The probability of not spinning red on one spin is the sum of the probabilities of landing on blue or cyan.

\[ P(\text{Not Red}) = P(\text{Blue}) + P(\text{Cyan}) \]

\[ P(\text{Blue}) = \frac{2}{6} = \frac{1}{3} \] \[ P(\text{Cyan}) = \frac{1}{6} \]

\[ P(\text{Not Red}) = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \]

Since the spins are independent, the probability of not spinning red on both spins is the product of the probabilities of not spinning red on each spin.

\[ P(\text{Not Red on Both Spins}) = P(\text{Not Red}) \times P(\text{Not Red}) \]

\[ P(\text{Not Red on Both Spins}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]

Final Answer