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. \(\frac{1}{32}\)

Answer 1

The Correct Option is D

We draw a marble 4 times with replacement from the set {1, 2, 5, 10}. Each draw is independent. Since there are 4 possible outcomes for each of the 4 draws, the total number of possible sequences (outcomes in the sample space) is:
Total Outcomes = 4 × 4 × 4 × 4 = 4⁴ = 256.

Identify Favorable Outcomes (Sum = 18)

We need to find sequences of 4 numbers (n₁, n₂, n₃, n₄) chosen from {1, 2, 5, 10} such that n₁ + n₂ + n₃ + n₄ = 18.

Let's find the combinations of numbers that sum to 18:

• The largest number is 10. If we use 10 twice (10+10=20), the sum is already > 18. So, we can use '10' at most once.
• If we don't use '10', the maximum sum is 5+5+5+5 = 20. Let's try combinations using only {1, 2, 5}:
  • 5, 5, 5, ? -> Sum is 15, need 3 (not in the set).
  • 5, 5, ?, ? -> Sum is 10, need 8 from two numbers in {1, 2, 5}. Possible pairs: (5, 3) - no 3; (2, 6) - no 6; (1, 7) - no 7. Cannot make 8 from {1, 2, 5}.
• So, we must use '10' exactly once. The remaining three numbers must sum to 18 - 10 = 8. These three numbers must be chosen from {1, 2, 5}.
  • Can we use 5? Yes. If one number is 5, the other two must sum to 8 - 5 = 3. From {1, 2, 5}, the only way to get 3 is 1 + 2. So, the three numbers are {5, 2, 1}.
  • Can we avoid using 5? The three numbers must be from {1, 2} and sum to 8. The maximum sum is 2+2+2 = 6. So, we cannot make 8 using only {1, 2}.

Therefore, the only combination of numbers (ignoring order for now) that sums to 18 is {1, 2, 5, 10}.

The favorable outcomes are the sequences formed by arranging the numbers {1, 2, 5, 10}. Since all four numbers are distinct, the number of ways to arrange them in a sequence of 4 draws is the number of permutations of these 4 numbers. Number of Favorable Sequences = 4! = 4 × 3 × 2 × 1 = 24.

The probability is the ratio of the number of favorable sequences to the total number of possible sequences. Probability = (Number of Favorable Sequences) / (Total Number of Outcomes) 
Probability = \(\frac {24}{256}\)

Simplify the fraction: Divide both numerator and denominator by their greatest common divisor, which is 8. 
Probability =\(\frac {(24 ÷ 8)}{(256 ÷ 8) }\)= \(\frac {3}{32}\).

The probability that the sum of the numbers equals 18 is 3/32.

The correct option is (D) : \(\frac{3}{32}\)

Answer 2

We have four marbles marked with {1, 2, 5, 10}. We select one marble four times with replacement and want to find the probability that the sum of the numbers equals 18.

Since we are selecting with replacement, there are \(4^4 = 256\) total possible outcomes.

We need to find combinations of the numbers {1, 2, 5, 10} that sum to 18, considering that we can select four numbers. Let's list the possibilities:

• 10 + 5 + 2 + 1 = 18. This combination uses all four numbers. There are 4! = 24 permutations.
• 10 + 2 + 2 + 2+2 10+2+6 no
• 10 + 5 +1+2 10,5,2,1=18
• 10 + 5 + 5 + - 2 = This one cannot be done to -2! we cannot change this

Since the elements must be in {1, 2, 5, 10} Let's find the possibilities:

10 + 5+ 2+1 =18 Total number of permutations 4! = 24 Since total outcomes = \(4^4 =256\) However, in the 256 outcomes, number of ways we can derive 10+5+2+1 Then 1/8 This method isnt working lets try the new one

• 1) 10+5+2+1 4! =24 permutations =24 possible combinations Here permutations include 4 numbers 1 selection
• 2) 5+5+5+3 = 0 valid case
• 3) since maximum number is 10: all other digits has to be lesser: no combination that exceeds digit number The greatest possibility will require largest number of singletons + combination number 4+3+2+1

Since it can have only 10,5,1,2 Thus 4! = 24 number permutation Possible combinations include 256 Thus 24/256 Then 24/(4*4*4*4) = \(\frac{3}{32}\)

• To verify this I must try a formula \(\binom{n+k-1}{k-1} \) THIS IS VALID ONLY IF THE COMBINATION is 1, 2, 3! for combination which has limit This does not fulfill the constraint. Thus number is 1 solution

 

Total permutations 4!/4!=24 total combinations are possible There are a total of = 256 ways one might select the numbers We can have permutation formula here Total combinations =4 thus \(24/256= 3/32\) This is the Answer!

There would be 3 cases of 18 in the digit 10,2,5,1 all of these would be equal each and each of this would have permutation of 4 ! permutation combinations to arrange digit combination 
 

Thus, probability to derive is 25 = Total Digit Combination which equals 3/32<4

 

Therefore, the probability is \(\frac{24}{256} = \frac{3}{32}\).