Question

A spinner is divided into a red section and a blue section. In a game, the spinner is spun twice. If the spinner lands on red at least once, then Kim wins the game; otherwise, Natalie wins the game. What fraction of the spinner must be red for the game to be fair? A. 1/4B. 1 - square root 2/2C. 1/2D. square root 2/2

Answer 1

A: the spinner lands on red

If you spin the spinner twice, the results are independent, then the probability of getting two reds is P(A)*P(A)

We need this probability equal to 1/2 in order to get a fair game. Since there are 2 possible outcomes (red or blue), then P(A) = x/2. Now we can write the next equation:

`\begin{gathered} (x)/(2)\cdot(x)/(2)=(1)/(2) \\ (x^2)/(4)=(1)/(2) \\ x^2=(4)/(2) \\ x=\sqrt[]{2} \end{gathered}`

This means that √2/2 of the spinner must be red