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:
We are conducting an experiment of drawing a marble 4 independent times, putting it back after each draw (with replacement).
The set of available numbers is \( \{1, 2, 5, 10\} \).
We need to find the theoretical probability of a specific event occurring: the sum of the four drawn numbers being exactly 18.
To do this, we compute the total possible outcomes and count how many of those outcomes result in a sum of 18.
Step 2: Key Formula or Approach:
1. Total Outcomes: If an experiment has \( n \) possible outcomes and is repeated \( k \) times with replacement, the total number of sequences is \( n^k \).
2. Permutations with Repetition: The number of distinct ways to arrange \( N \) items where there are duplicates (like \( n_1 \) of one kind, \( n_2 \) of another) is \( \frac{N!}{n_1! n_2! \dots} \).
3. Probability: \( P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \).
Step 3: Detailed Explanation:
Let's find the total number of outcomes in the sample space.
Since there are 4 marbles and we draw 4 times with replacement, each draw has 4 possibilities.
\[ \text{Total Outcomes} = 4 \times 4 \times 4 \times 4 = 4^4 = 256 \] Next, we must find the favorable combinations of four numbers from the set \( \{1, 2, 5, 10\} \) that sum to exactly 18.
Let the drawn numbers be \( x_1, x_2, x_3, x_4 \). We want \( x_1 + x_2 + x_3 + x_4 = 18 \).
We analyze this by considering how many times the largest number, 10, can be drawn.
Case 1: The number 10 is drawn twice or more.
If we draw two 10s, the sum is already \( 10 + 10 = 20 \), which is strictly greater than 18. This is impossible.
Case 2: The number 10 is drawn exactly once.
If one draw is a 10, the remaining three draws must sum to \( 18 - 10 = 8 \).
We need three numbers from \( \{1, 2, 5\} \) that add to 8.
Let's test using the next largest number, 5.
If we use one 5, the remaining two numbers must sum to \( 8 - 5 = 3 \).
From \( \{1, 2, 5\} \), the only two numbers that sum to 3 are 1 and 2.
So, a valid set of four numbers is \( \{10, 5, 2, 1\} \).
Are there other ways to make 8?
- If we use two 5s, the sum is 10 (too large).
- If we use zero 5s, the maximum sum with three numbers from \( \{1, 2\} \) is \( 2 + 2 + 2 = 6 \) (too small).
Thus, the only specific set of numbers that works is \( \{10, 5, 2, 1\} \).
Because all four drawn numbers are distinct, any permutation of these 4 numbers represents a different valid sequence of draws.
The number of distinct arrangements of 4 unique items is \( 4! \).
\[ \text{Favorable outcomes from Case 2} = 4! = 4 \times 3 \times 2 \times 1 = 24 \] Case 3: The number 10 is drawn zero times.
We must make a sum of 18 using four numbers strictly from \( \{1, 2, 5\} \).
The maximum possible sum using four numbers from this restricted set is by picking the largest number, 5, four times.
Max sum = \( 5 + 5 + 5 + 5 = 20 \).
We need to reduce this sum from 20 down to 18.
If we replace one '5' with a '2', the sum drops to \( 5 + 5 + 5 + 2 = 17 \). This is strictly less than 18.
Since replacing even a single 5 with the next largest number makes the sum fall below 18, it is completely impossible to get a sum of 18 without using the number 10.
Final Calculation:
The total number of favorable outcomes is exclusively the 24 outcomes found in Case 2.
\[ \text{Probability} = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{24}{256} \] Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 8:
\[ \frac{24 \div 8}{256 \div 8} = \frac{3}{32} \] Step 4: Final Answer:
The probability is \( \frac{3}{32} \).