A die is loaded in such a way that the probability of gettin k (k=1,2,3,...6) on the die is propotional to K. What is the probability of getting a product of 6n; n belong to natural number if it is rolled thrice

Given that the probability of getting a face value \( k \) on a loaded die is proportional to \( k \), the probability of each face value can be expressed as follows:

\[
P(k) = \frac{k}{\sum_{k=1}^{6} k} = \frac{k}{21}
\]

where \( k = 1, 2, 3, 4, 5, 6 \), and the sum of the numbers from 1 to 6 is:

\[
\sum_{k=1}^{6} k = 1 + 2 + 3 + 4 + 5 + 6 = 21
\]

Thus, the probability distribution for each face value is:

\[
P(1) = \frac{1}{21}, P(2) = \frac{2}{21}, P(3) = \frac{3}{21}, P(4) = \frac{4}{21}, P(5) = \frac{5}{21}, P(6) = \frac{6}{21}
\]

### Goal:
We need to find the probability that the product of the values obtained when rolling the die three times equals \( 6n \), where \( n \) is a natural number.

#### Steps:

1. **Possible Outcomes that Yield Products of 6n:**
   The factors of 6 are \( 2 \times 3 \). Hence, any product of the three numbers rolled must include a combination of \( 2 \), \( 3 \), and possibly other factors. To get \( 6n \), the product of the three rolls must be divisible by 6, which implies the outcome must include both a 2 and a 3 among the three dice.

2. **Breakdown of Required Combinations:**
   To have a product divisible by 6, you need:
   - At least one die to show a value of 2 (or any multiple of 2, i.e., 2, 4, or 6).
   - At least one die to show a value of 3 (or any multiple of 3, i.e., 3 or 6).
   
   However, the product of all three numbers does not have to exceed this. For instance, rolls like \( (2, 3, 1) \), \( (4, 3, 1) \), \( (6, 1, 1) \), etc., will yield a product divisible by 6.

3. **Calculating Probabilities:**
   We can count the favorable outcomes and find the probability by considering all rolls that meet the condition of the product being divisible by 6. Let me compute the result.

The probability that the product of the values obtained when rolling a loaded die three times is divisible by 6 is approximately \( 0.758 \), or 75.8%.

Would you like more details or have any other questions?

Here are 5 related questions you might find interesting:
1. What would the probability be if the die were fair?
2. How would the probability change if the die were rolled four times?
3. What if we want the product to be exactly 36 instead of any multiple of 6?
4. How can we calculate the probability of obtaining a sum divisible by 6 instead of a product?
5. How would the answer change if the probabilities of the die were reversed (e.g., lower numbers more probable)?

**Tip:** Whenever solving probability problems with multiple rolls, it's helpful to start by identifying key factors or conditions the outcomes must satisfy, as it narrows down the possible cases.

A die is loaded in such a way that the probability of getting k (k=1,2,3,...6) on the die is proportional to K. What is the probability of getting a product of 6n; n belong to natural number if it is rolled thrice?

Explore the probability of getting a product divisible by 6 when rolling a loaded die three times. Learn how the face values are proportional to their outcomes and how divisibility rules for 6 apply to probability. Detailed breakdown with key formulas and concepts for Grades 9-12.

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