The problem asks for the probability of spinning a spinner twice and landing on a number greater than 5 on the first spin, and then landing on a number less than 5 on the second spin. The answer should be expressed as a percentage. The spinner has 4 equally sized sections with the numbers 3, 4, 5, and 6.

Probability and StatisticsProbabilityIndependent EventsSpinnerPercentage
2025/3/13

1. Problem Description

The problem asks for the probability of spinning a spinner twice and landing on a number greater than 5 on the first spin, and then landing on a number less than 5 on the second spin. The answer should be expressed as a percentage. The spinner has 4 equally sized sections with the numbers 3, 4, 5, and
6.

2. Solution Steps

First, we need to determine the probability of landing on a number greater than 5 on the first spin. The only number greater than 5 on the spinner is

6. Since there are 4 equally sized sections, the probability of landing on 6 is $\frac{1}{4}$.

Next, we need to determine the probability of landing on a number less than 5 on the second spin. The numbers less than 5 are 3 and

4. Since there are 4 equally sized sections, the probability of landing on 3 or 4 is $\frac{2}{4} = \frac{1}{2}$.

Since the two spins are independent events, we multiply their probabilities to find the probability of both events occurring.
P(greater than 5 and then less than 5)=P(greater than 5)×P(less than 5)P(\text{greater than 5 and then less than 5}) = P(\text{greater than 5}) \times P(\text{less than 5})
P(greater than 5 and then less than 5)=14×12=18P(\text{greater than 5 and then less than 5}) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}
Now, we convert the fraction to a percentage.
18=0.125\frac{1}{8} = 0.125
0.125×100=12.50.125 \times 100 = 12.5

3. Final Answer

1

2. 5%

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