sum equal to S, then:
There are 2 six faced dice, one of them is Normal and the other is loaded in surf a way that the probability of getting a number is proportional to the number itself. The two dice are thrown together. If P(S) is the event of getting a
Pick ONE option
P(S≥11)=11/126
P(S≥11)=17/126
P(S≥11)=19/126
None of the above

We have two six-faced dice, one normal and the other loaded such that the probability of getting a number on the loaded die is proportional to the number itself. Let’s break down how to calculate the probability \( P(S \geq 11) \), where \( S \) is the sum of the numbers on the two dice when they are thrown together.

### Step 1: Normal die probability
For the normal die, the probability of rolling any number between 1 and 6 is uniformly distributed, i.e., each outcome has a probability of:
\[
P(\text{normal die} = k) = \frac{1}{6}, \quad k = 1, 2, 3, 4, 5, 6.
\]

### Step 2: Loaded die probability
For the loaded die, the probability of rolling a number is proportional to the number itself. Hence, the probabilities of getting 1, 2, 3, 4, 5, or 6 are proportional to 1, 2, 3, 4, 5, and 6, respectively. The sum of these weights is:
\[
1 + 2 + 3 + 4 + 5 + 6 = 21.
\]
Thus, the probability of rolling a \( k \) on the loaded die is:
\[
P(\text{loaded die} = k) = \frac{k}{21}, \quad k = 1, 2, 3, 4, 5, 6.
\]

### Step 3: Find the probability of \( S \geq 11 \)
Now we compute the probability of obtaining a sum \( S \geq 11 \), where \( S = x + y \) and \( x \) is the outcome of the normal die and \( y \) is the outcome of the loaded die. The possible sums that are \( \geq 11 \) are 11 and 12.

#### Case 1: \( S = 11 \)
The pairs \( (x, y) \) that sum to 11 are:
- \( (5, 6) \): Probability \( P(5, 6) = \frac{1}{6} \times \frac{6}{21} = \frac{1}{21} \),
- \( (6, 5) \): Probability \( P(6, 5) = \frac{1}{6} \times \frac{5}{21} = \frac{5}{126} \).

So, the total probability for \( S = 11 \) is:
\[
P(S = 11) = \frac{1}{21} + \frac{5}{126} = \frac{6}{126} + \frac{5}{126} = \frac{11}{126}.
\]

#### Case 2: \( S = 12 \)
The only pair \( (x, y) \) that sums to 12 is \( (6, 6) \), with a probability:
\[
P(6, 6) = \frac{1}{6} \times \frac{6}{21} = \frac{1}{21}.
\]

Thus, the total probability for \( S \geq 11 \) is:
\[
P(S \geq 11) = P(S = 11) + P(S = 12) = \frac{11}{126} + \frac{6}{126} = \frac{17}{126}.
\]

### Conclusion
The correct probability is:
\[
P(S \geq 11) = \frac{17}{126}.
\]

Thus, the correct answer is:
\[
\boxed{P(S \geq 11) = \frac{17}{126}}.
\]

Would you like further clarification or details on any part of this solution?

Here are 5 questions related to this problem:
1. How do probabilities on a loaded die differ from a normal die?
2. How would the probability change if both dice were loaded?
3. What is the probability of obtaining \( S = 7 \) with these two dice?
4. How would the result change if the loaded die favored smaller numbers?
5. Can the outcome probabilities be represented using a probability distribution function?

**Tip:** When dealing with a loaded die, remember that the probability of each face is not uniform but weighted according to the given distribution.

There are 2 six-faced dice, one of them is normal and the other is loaded in such a way that the probability of getting a number is proportional to the number itself. The two dice are thrown together. If P(S) is the event of getting a sum equal to or greater than S, what is the correct probability? P(S≥11) = 11/126, 17/126, 19/126, or None of the above.

Explore how to calculate the probability of obtaining a sum of 11 or greater when rolling a normal die and a loaded die, where the loaded die's probabilities are weighted. This step-by-step guide breaks down the formula and theorems used, perfect for high school students studying advanced probability.

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