Question

the fair spinner shown is spun 2 times. the set of possible outcomes are: s = {rr, bb, yy, gg, rb, br, ry, yr, rg, gr, by, yb, bg, gb, yg, gy} the number of yellows in the outcome is summarized in the table on the right (yellow: 0, 1, 2; frequency: 9, 6, 1). using the data from the theoretical probability table, what is the probability of the spinner landing only once on yellow in two spins? options: 0.0625, 0.5625, 0.375, 1

Explanation:

Step1: Determine total outcomes

The spinner has 4 equal sections (let's assume colors: R, B, Y, G). When spun twice, total outcomes \( n(S)=4\times4 = 16 \).

Step2: Identify favorable outcomes (1 yellow)

Favorable outcomes: (RY, YR, BY, YB, GY, YG). Count: 6. Wait, no—wait, the table: Yellow 0:9, 1:6, 2:1. Wait, theoretical probability: each spin, P(Yellow) = \( \frac{1}{4} \), P(not Yellow) = \( \frac{3}{4} \). For 1 yellow in two spins, use binomial probability: \( P(X = 1)=\binom{2}{1}\times(\frac{1}{4})^1\times(\frac{3}{4})^1 \).

Step3: Calculate binomial probability

\( \binom{2}{1}=\frac{2!}{1!(2 - 1)!}=2 \). So \( P = 2\times\frac{1}{4}\times\frac{3}{4}=\frac{6}{16}=0.375 \).

Answer:

0.375