Question

Three numbers are chosen at random from 1 to 20. The probability that their sum is divisible by 3 is:

• A. \( \frac{1}{114} \)
• B. \( \frac{147}{342} \)
• C. \( \frac{16}{47} \)
• D. \( \frac{32}{85} \)

Hint:

To check divisibility by 3 when selecting numbers randomly, classify numbers based on their remainder when divided by 3 and count valid cases.

Answer 1

The Correct Option is D

We need to calculate the probability that the sum of three randomly chosen numbers from 1 to 20 is divisible by 3. 

Step 1: Understanding Residues Modulo 3 The numbers from 1 to 20 have residues (remainders) when divided by 3: 
- Numbers with remainder 0 (divisible by 3): {3, 6, 9, 12, 15, 18} 
⇒ 6 numbers - Numbers with remainder 1: {1, 4, 7, 10, 13, 16, 19} 
⇒ 7 numbers - Numbers with remainder 2: {2, 5, 8, 11, 14, 17, 20} ⇒ 7 numbers 

Step 2: Conditions for Sum to be Divisible by 3 For the sum of three numbers to be divisible by 3, we need one of the following combinations: - All three numbers have the same remainder.
 - One number from each residue group. 

Step 3: Counting Possible Combinations 
Case 1: All numbers have the same remainder 
- All zero remainders ⇒ \( \binom{6}{3} = 20 \) - All one remainders
 ⇒ \( \binom{7}{3} = 35 \) - All two remainders
 ⇒ \( \binom{7}{3} = 35 \) 
Total combinations for this case: \[ 20 + 35 + 35 = 90 \] 
Case 2: One number from each residue group Choosing one from each category: \[ 6 \times 7 \times 7 = 294 \] 

Step 4: Total Favorable Outcomes \[ \text{Total favorable outcomes} = 90 + 294 = 384 \] 

Step 5: Total Possible Outcomes The total number of ways to choose 3 numbers from 20 is: \[ \binom{20}{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140 \] 

Step 6: Probability Calculation \[ \text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{384}{1140} = \frac{32}{95} \] Step 7: Final Answer 

\[Correct Answer: (4) \ \frac{32}{85}\]

Answer 2

To solve the problem of finding the probability that the sum of three randomly chosen numbers from 1 to 20 is divisible by 3, we begin by considering modulo arithmetic.

Firstly, we classify the numbers from 1 to 20 based on their remainder when divided by 3:

• Numbers with remainder 0: 3, 6, 9, 12, 15, 18 (6 numbers)
• Numbers with remainder 1: 1, 4, 7, 10, 13, 16, 19 (7 numbers)
• Numbers with remainder 2: 2, 5, 8, 11, 14, 17, 20 (7 numbers)

The sum of three numbers will be divisible by 3 if their remainders sum to a multiple of 3. Let us consider the cases:

1. All selected numbers are from the same remainder group: This means choosing 3 numbers with remainder 0, 1, or 2.
2. One number from each remainder group: This allows a combination of remainders 0, 1, and 2.

Calculating each case:

• All numbers with remainder 0: Choose 3 numbers from the 6 available. There are $\frac{6C3}{}$ = 20 ways.
• All numbers with remainder 1: Choose 3 numbers from the 7 available. There are $\frac{7C3}{}$ = 35 ways.
• All numbers with remainder 2: Choose 3 numbers from the 7 available. There are $\frac{7C3}{}$ = 35 ways.
• One number from each group: Choose one number from each group (6 from group 0, 7 from group 1, 7 from group 2). The number of ways = 6 * 7 * 7 = 294.

Total favorable outcomes: \(20 + 35 + 35 + 294 = 384\)

Total possible outcomes: Choosing any 3 numbers from 20 without restrictions, which is $\frac{20C3}{}$ = 1140.

Probability: \(\frac{384}{1140} = \frac{32}{95}\)