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a spinner contains 3 equal sections: one red, one yellow, and one blue.…

Question

a spinner contains 3 equal sections: one red, one yellow, and one blue. the possible results of two spins are shown in the tree diagram below. what is p(red, then yellow)? \\(\frac{1}{9}\\) \\(\frac{2}{9}\\) \\(\frac{1}{3}\\) \\(\frac{2}{3}\\) \
tree diagram: first spin branches: r, y, b. from r: second spin r (r,r), y (r,y), b (r,b). from y: second spin r (y,r), y (y,y), b (y,b). from b: second spin r (b,r), y (b,y), b (b,b).

Explanation:

Step1: Find P(red first)

The spinner has 3 equal sections, so the probability of landing on red first is $\frac{1}{3}$ (since there's 1 red out of 3 sections).

Step2: Find P(yellow second | red first)

After spinning red first, the spinner is still 3 equal sections, so the probability of landing on yellow next is also $\frac{1}{3}$ (1 yellow out of 3 sections).

Step3: Multiply the probabilities

For independent events (spins are independent), $P(\text{red, then yellow}) = P(\text{red}) \times P(\text{yellow} | \text{red})$. So $\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$. Alternatively, count the outcomes: total outcomes are 9 (from the tree diagram: 3 first spins × 3 second spins). The outcome (R, Y) is 1, so probability is $\frac{1}{9}$.

Answer:

$\frac{1}{9}$ (corresponding to the first option $\frac{1}{9}$)