Question

A four-digit number is formed using the digits 1, 2, 3, 4, 5 without repetition. The probability that is divisible by 3 is

• A. \( \frac{1}{3} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{1}{5} \)
• D. \( \frac{1}{6} \)

Hint:

Remember the divisibility rule for 3: a number is divisible by 3 if and only if the sum of its digits is divisible by 3. In selection problems, first find the sum of all available digits. Then, determine which digits must be included or excluded to make the sum of the selected digits divisible by 3.

Answer 1

The Correct Option is C

Step 1: Find the total number of possible outcomes.
We need to form a four-digit number using the digits {1, 2, 3, 4, 5} without repetition.
The number of ways to choose and arrange 4 digits from 5 is given by the permutation formula \( ^5P_4 \).
Total outcomes = \( ^5P_4 = \frac{5!}{(5-4)!} = 5! = 120 \).

Step 2: Find the number of favorable outcomes.
A number is divisible by 3 if the sum of its digits is divisible by 3.
The sum of all five available digits is \( 1 + 2 + 3 + 4 + 5 = 15 \), which is divisible by 3.
To form a four-digit number, we must leave out one digit. For the sum of the remaining four digits to be divisible by 3, the digit we leave out must also be divisible by 3.
From the set {1, 2, 3, 4, 5}, the only digit divisible by 3 is 3. So, to form a number divisible by 3, we must use the digits {1, 2, 4, 5}.
The sum is \( 1+2+4+5=12 \), which is divisible by 3.
Alternatively, if we remove 1, sum is 14 (not div by 3). Remove 2, sum is 13 (not div by 3). Remove 4, sum is 11 (not div by 3). Remove 5, sum is 10 (not div by 3).
So, the only combination of four digits that works is {1, 2, 4, 5}. The number of four-digit numbers that can be formed using these four digits is \( 4! \).
Favorable outcomes = \( 4! = 4 \times 3 \times 2 \times 1 = 24 \).

Step 3: Calculate the probability. Probability = \( \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{24}{120} \). \[ \frac{24}{120} = \frac{1}{5} \]