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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. | yellow | frequency | | --- | --- | | 0 | 9 | | 1 | 6 | | 2 | 1 | using the data from the theoretical probability table, what is the probability of the spinner landing only once on yellow in two spins? 0.375 0.5625 0.0625 1
Step1: Find total number of outcomes
First, we calculate the total number of possible outcomes by summing up all the frequencies. The frequencies are 9 (for 0 yellows), 6 (for 1 yellow), and 1 (for 2 yellows). So total outcomes \( n = 9 + 6 + 1 = 16 \).
Step2: Find frequency for 1 yellow
The frequency of the spinner landing once on yellow (i.e., number of yellows = 1) is given as 6.
Step3: Calculate probability
Probability \( P \) of an event is the number of favorable outcomes divided by the total number of outcomes. So, \( P(\text{1 yellow}) = \frac{\text{Frequency of 1 yellow}}{\text{Total number of outcomes}} = \frac{6}{16} = 0.375 \).
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0.375