Question

riley is playing a game that requires rolling a number cube numbered 1–6 and spinning a spinner with 3 blue sections and 1 green section. the sections of the spinner are all the same size. what is the probability that riley rolls a number less than 3 and lands on a blue section? \frac{1}{2} \frac{1}{8} \frac{1}{4} \frac{3}{4}

Explanation:

Step1: Find probability of rolling less than 3

A number cube (1 - 6) has numbers less than 3: 1, 2. So there are 2 favorable outcomes. Total outcomes = 6. Probability \( P(\text{roll} < 3) = \frac{2}{6} = \frac{1}{3} \)? Wait, no, 1 and 2 are two numbers, so \( \frac{2}{6}=\frac{1}{3} \)? Wait, no, 1,2: two numbers, total 6. So \( P(\text{roll} < 3) = \frac{2}{6} = \frac{1}{3} \)? Wait, no, wait: numbers less than 3 on 1 - 6 are 1 and 2, so 2 outcomes. So \( \frac{2}{6} = \frac{1}{3} \). Wait, no, maybe I miscalculated. Wait, 1,2: two numbers, so probability is \( \frac{2}{6} = \frac{1}{3} \). Then, spinner has 3 blue and 1 green, total 4 sections. Probability of blue: \( \frac{3}{4} \). Now, since rolling and spinning are independent, multiply the probabilities: \( \frac{1}{3} \times \frac{3}{4} = \frac{1}{4} \). Wait, no, wait: wait, numbers less than 3: 1 and 2, so 2 out of 6, which is \( \frac{2}{6} = \frac{1}{3} \). Spinner: 3 blue out of 4, so \( \frac{3}{4} \). Then, independent events, so multiply: \( \frac{1}{3} \times \frac{3}{4} = \frac{1}{4} \). Wait, but let's check again. Wait, maybe I made a mistake in the first probability. Wait, numbers less than 3: 1,2. So 2 numbers. Total 6. So \( P(\text{roll} < 3) = \frac{2}{6} = \frac{1}{3} \). Spinner: 3 blue, 1 green, total 4. So \( P(\text{blue}) = \frac{3}{4} \). Then, the combined probability (since the events are independent) is \( \frac{1}{3} \times \frac{3}{4} = \frac{1}{4} \). So the probability is \( \frac{1}{4} \), which is one of the options (the third option, \( \frac{1}{4} \)). Wait, let's re - do the first step. Wait, numbers less than 3 on a die (1 - 6) are 1 and 2, so 2 favorable outcomes. Total outcomes: 6. So \( P(\text{roll} < 3)=\frac{2}{6}=\frac{1}{3} \). Spinner: 3 blue sections, 1 green section, total 4 sections. So \( P(\text{blue})=\frac{3}{4} \). Since the two events (rolling the die and spinning the spinner) are independent, we multiply the probabilities: \( P(\text{roll}<3 \text{ and blue}) = P(\text{roll}<3)\times P(\text{blue})=\frac{2}{6}\times\frac{3}{4}=\frac{2\times3}{6\times4}=\frac{6}{24}=\frac{1}{4} \).

Answer:

\(\frac{1}{4}\) (the third option, e.g., if the options are labeled as first: \(\frac{1}{2}\), second: \(\frac{1}{8}\), third: \(\frac{1}{4}\), fourth: \(\frac{3}{4}\), then the answer is the third option, \(\frac{1}{4}\))