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.