Question

In a box there are four marbles and each of them is marked with distinct number from the set \( \{1, 2, 5, 10\} \). If one marble is randomly selected four times with replacement and the number on it noted, then the probability that the sum of numbers equals 18 is:

• A. \(\frac{1}{64}\)
• B. \(\frac{3}{16}\)
• C. \(\frac{5}{32}\)
• D. \(\frac{3}{32}\)
• E. \(\frac{1}{32}\)

Hint:

When the set of numbers is small, listing combinations systematically (starting with the largest possible value) is often more reliable than complex generating functions.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
This is a probability problem involving a discrete sample space where each selection is independent due to replacement. We need to find the number of ways four numbers chosen from $\{1, 2, 5, 10\}$ can sum to 18, and divide by the total number of possible outcomes.

Step 2: Key Formula or Approach
1. Total outcomes = \(4 \times 4 \times 4 \times 4 = 4^4 = 256\).
2. Find combinations of \(\{x_1, x_2, x_3, x_4\}\) such that \(\sum x_i = 18\).

Step 3: Detailed Explanation
We look for combinations of four numbers from \(\{1, 2, 5, 10\}\) that sum to 18:
• Case 1: Using two 10s. The remaining two numbers must sum to \(-2\), which is impossible.
• Case 2: Using one 10. The remaining three numbers must sum to 8.
• \((5, 2, 1)\): Sum is 8. Permutations: \(4! = 24\) ways.
• \((2, 2, 4)\): Not possible (4 is not in the set).
• \((10, 5, 2, 1)\): The permutations are \(4! = 24\).
• Case 3: Using zero 10s. We must use 5s, 2s, and 1s to reach 18.
• \((5, 5, 5, 3)\): Not possible.
• \((5, 5, 6)\): Not possible.
• Let's check \((5, 5, 5, \dots)\): \(15 + x = 18 \implies x=3\) (No).
• Let's check \((5, 5, 5, 2, 1)\): This is 5 numbers, we need 4.
• Let's check \((5, 5, \dots)\): \(10 + x + y = 18 \implies x+y=8\). Possible pair is \((?)\). Only \(5+2+1\) was 8. So \((5, 5, 5, \dots)\) fails.
• Wait, check \((10, 2, 5, 1)\) again. That is one 10, one 5, one 2, one 1. Sum = 18. Ways = 24.
• Are there other combinations for sum 8? \((2, 2, 2, 2)\) is 8. So \((10, 2, 2, 2, 2)\) is 5 numbers.
• Let's check for sum 18 again:
• \((10, 2, 5, 1)\) $\rightarrow$ 24 ways.
• \((5, 5, 5, \dots)\) $\rightarrow$ No.
• \((10, 5, 1, 2)\) $\rightarrow$ Already counted.
• \((10, 2, 2, ?)\) $\rightarrow$ No.
• Is there any other? How about \((5, 5, 5, ?)\) $\rightarrow$ No.
• How about \((5, 5, 2, ?)\) $\rightarrow$ \(10 + 2 = 12\), need 6 more. No.
• Let's look at \((10, \dots)\) again. \(10 + 5 + 1 + 2 = 18\).
• Let's look at \((?)\). Oh, \((10, 10, \dots)\) was impossible.
• Let's check \((5, 5, 5, ?)\) - No.
• What about \((5, 5, 2, ?)\) - No.
• What about \((5, 5, 5, ?)\) - No.
• Let's re-examine: One 10, one 5, one 2, one 1. Sum = 18. (24 ways).
• Are there any others? \((5, 5, 5, 5)\) is 20. \((5, 5, 5, 2)\) is 17. \((5, 5, 5, 1)\) is 16.
• \((10, 2, 2, 2)\) is 16. \((10, 5, 5, ?)\) is 20.
• The only combination is \((10, 5, 2, 1)\). Probability = \(24/256 = 3/32\).

Step 4: Final Answer
The probability is \(3/32\).