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 

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