3-digit numbers are formed using the digits 1, 3, 7 without repetition of digits. A number is randomly selected. What is the probability that the number is divisible by 3?
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Understanding Probability with 3-Digit Numbers
The question asks for the probability that a randomly selected 3-digit number, formed using the digits 1, 3, and 7 without repetition, is divisible by 3.
Step 1: Determine the Total Number of Possible Outcomes (Sample Space)
We are forming 3-digit numbers using the digits 1, 3, and 7 without repetition. The total number of distinct 3-digit numbers that can be formed is the number of permutations of 3 distinct items taken 3 at a time, denoted as \(P(3,3)\) or \(3!\).
Calculation:
\(P(3,3) = 3! = 3 \times 2 \times 1 = 6\)
The 6 possible numbers are: 137, 173, 317, 371, 713, 731.
So, the total number of outcomes in our sample space is 6.
Step 2: Determine the Number of Favorable Outcomes (Numbers Divisible by 3)
A key rule for divisibility by 3 is that a number is divisible by 3 if and only if the sum of its digits is divisible by 3.
Let's find the sum of the digits used: 1, 3, and 7.
Sum of digits = \(1 + 3 + 7 = 11\).
Now, check if this sum (11) is divisible by 3. \(11 \div 3\) gives a remainder (11 = 3 \times 3 + 2). So, 11 is not divisible by 3.
Since the sum of the digits (11) is not divisible by 3, any number formed by rearranging these specific digits (1, 3, and 7) will also not be divisible by 3. This holds true for all the 6 possible numbers we listed in Step 1.
Therefore, the number of favorable outcomes (numbers divisible by 3) is 0.
Step 3: Calculate the Probability
The probability of an event is calculated as:
Probability = (Number of favorable outcomes) / (Total number of possible outcomes)
In this case:
Probability (Number is divisible by 3) = (Number of 3-digit numbers divisible by 3) / (Total number of 3-digit numbers formed)
Probability = \(0 / 6\)
Probability = \(0\)
Thus, the probability that the randomly selected 3-digit number formed using digits 1, 3, and 7 without repetition is divisible by 3 is 0.
Summary of Steps
- Calculate the total possible 3-digit numbers using 1, 3, 7 without repetition (Permutations).
- Check the divisibility rule for 3 based on the sum of digits.
- Determine how many formed numbers satisfy the divisibility rule.
- Calculate probability: (Favorable Outcomes) / (Total Outcomes).
| Concept | Value/Outcome |
|---|---|
| Digits Used | 1, 3, 7 |
| Constraint | No repetition |
| Total 3-Digit Numbers (Sample Space) | \(P(3,3) = 6\) |
| Sum of Digits | \(1 + 3 + 7 = 11\) |
| Is Sum Divisible by 3? | No (11 is not divisible by 3) |
| Number of Numbers Divisible by 3 (Favorable Outcomes) | 0 |
| Probability | \(0/6 = 0\) |
Revision Table: Probability and Divisibility
| Term | Definition/Concept |
|---|---|
| Probability | A measure of the likelihood of an event occurring. Calculated as (Favorable Outcomes) / (Total Outcomes). |
| Sample Space | The set of all possible outcomes in a probability experiment. |
| Favorable Outcome | An outcome that meets the specific condition or event being considered. |
| Divisibility Rule for 3 | A number is divisible by 3 if the sum of its digits is divisible by 3. |
| Permutation | An arrangement of objects in a specific order. The number of permutations of \(n\) objects taken \(r\) at a time is \(P(n,r) = n! / (n-r)!\). When \(n=r\), \(P(n,n) = n!\). |
Additional Information: Divisibility Rules Explained
Understanding divisibility rules can simplify problems involving factors and multiples. Here are a few common ones:
- Divisibility by 2: A number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8).
- Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3.
- Divisibility by 4: A number is divisible by 4 if the number formed by its last two digits is divisible by 4.
- Divisibility by 5: A number is divisible by 5 if its last digit is 0 or 5.
- Divisibility by 6: A number is divisible by 6 if it is divisible by both 2 and 3.
- Divisibility by 9: A number is divisible by 9 if the sum of its digits is divisible by 9. (Note: If a number is divisible by 9, it is also divisible by 3, but the converse is not always true).
- Divisibility by 10: A number is divisible by 10 if its last digit is 0.
In this problem, the sum of the digits 1, 3, and 7 is 11. Since 11 is not divisible by 3, any combination of these digits will form a number not divisible by 3. Thus, there are no favorable outcomes, resulting in a probability of 0.
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