"An ordinary fair die is thrown repeatedly until a 6 is obtained. Find the probability that it takes at least 3 throws but no more than 5 throws to obtain a 1 or  6."

To solve this problem, we'll break it down step by step:

### Step 1: Define the Situation
- We are rolling a fair die repeatedly until a 6 is obtained.
- We want to find the probability that it takes at least 3 throws but no more than 5 throws to obtain a 1 **or** a 6.

### Step 2: Identify the Probabilities
- The probability of rolling a 1 or a 6 on any single throw is \( \frac{2}{6} = \frac{1}{3} \).
- The probability of **not** rolling a 1 or a 6 (i.e., rolling a 2, 3, 4, or 5) on any single throw is \( \frac{4}{6} = \frac{2}{3} \).

### Step 3: Set Up the Events
We need to calculate the probability of getting either a 1 or a 6 on the 3rd, 4th, or 5th throw, with no such result before.

- **At least 3 throws** means that for the first two throws, neither a 1 nor a 6 is rolled. The probability for each of these throws is \( \frac{2}{3} \).
- The final throw (3rd, 4th, or 5th) must result in either a 1 or a 6, with probability \( \frac{1}{3} \).

### Step 4: Calculate the Desired Probability
We calculate the probabilities for each of the events where the 1st and 2nd rolls result in a non-1/6 outcome, followed by a success on the 3rd, 4th, or 5th roll.

1. **Probability of success on the 3rd throw**:
   \[
   P(\text{Success on 3rd throw}) = \left(\frac{2}{3}\right) \times \left(\frac{2}{3}\right) \times \left(\frac{1}{3}\right)
   \]
2. **Probability of success on the 4th throw**:
   \[
   P(\text{Success on 4th throw}) = \left(\frac{2}{3}\right) \times \left(\frac{2}{3}\right) \times \left(\frac{2}{3}\right) \times \left(\frac{1}{3}\right)
   \]
3. **Probability of success on the 5th throw**:
   \[
   P(\text{Success on 5th throw}) = \left(\frac{2}{3}\right) \times \left(\frac{2}{3}\right) \times \left(\frac{2}{3}\right) \times \left(\frac{2}{3}\right) \times \left(\frac{1}{3}\right)
   \]

### Step 5: Add the Probabilities Together
The total probability is the sum of the individual probabilities:

\[
P(\text{At least 3, no more than 5 throws}) = P(\text{Success on 3rd throw}) + P(\text{Success on 4th throw}) + P(\text{Success on 5th throw})
\]

Now, let's calculate this.The probability that it takes at least 3 throws but no more than 5 throws to obtain a 1 or a 6 is approximately **0.313** or 31.3%.

Would you like further details on the solution or have any other questions?

Here are 5 related questions to further explore:

1. How would the probability change if we were interested in at least 4 throws but no more than 6 throws?
2. What is the expected number of throws needed to get a 1 or a 6?
3. How would the probability differ if the die were biased, favoring some outcomes over others?
4. What is the probability that a 1 or a 6 is rolled on the first throw?
5. How would the total probability change if we consider two dice instead of one?

**Tip:** When solving probability problems involving multiple independent events, it is helpful to first break down the individual probabilities and then combine them using addition or multiplication rules.

An ordinary fair die is thrown repeatedly until a 6 is obtained. Find the probability that it takes at least 3 throws but no more than 5 throws to obtain a 1 or 6.

Explore the probability of rolling a 1 or 6 on a fair die within 3 to 5 throws. This problem demonstrates key concepts in probability, including the multiplication and addition rules, suitable for students in Grades 10-12.

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