Question:
Five digit numbers are formed by using digits \(1,2,3,4\) and \(5\) without repetition. Then, the probability that the randomly chosen number is divisible by \(4\) is
Five digit numbers are formed by using digits \(1,2,3,4\) and \(5\) without repetition. Then, the probability that the randomly chosen number is divisible by \(4\) is
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For divisibility by \(4\), always check only the last two digits of the number.
Updated On: Jun 22, 2026
- \(\dfrac{1}{5}\)
- \(\dfrac{5}{6}\)
- \(\dfrac{4}{5}\)
- \(\dfrac{1}{6}\)
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The Correct Option is A
Solution and Explanation
Step 1: Find the total number of five digit numbers.
The digits available are \(1,2,3,4,5\).
Since repetition is not allowed, the total number of five digit numbers formed using all five digits is
\[ 5! = 120 \]
Step 2: Use the divisibility rule of \(4\).
A number is divisible by \(4\) if its last two digits form a number divisible by \(4\).
So, we need to check the possible two digit endings using the digits \(1,2,3,4,5\).
Step 3: Find possible last two digits divisible by \(4\).
The two digit numbers formed from the given digits without repetition and divisible by \(4\) are
\[ 12,\;24,\;32,\;52 \] Thus, there are \(4\) possible choices for the last two digits.
Step 4: Arrange the remaining digits.
After fixing the last two digits, the remaining three digits can be arranged in
\[ 3! = 6 \] ways.
Therefore, the number of favorable five digit numbers is
\[ 4 \times 3! = 4 \times 6 = 24 \]
Step 5: Calculate the required probability.
\[ P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \] \[ P=\frac{24}{120} \] \[ P=\frac{1}{5} \]
Step 6: Final conclusion.
Hence, the required probability is
\[ \boxed{\frac{1}{5}} \]
The digits available are \(1,2,3,4,5\).
Since repetition is not allowed, the total number of five digit numbers formed using all five digits is
\[ 5! = 120 \]
Step 2: Use the divisibility rule of \(4\).
A number is divisible by \(4\) if its last two digits form a number divisible by \(4\).
So, we need to check the possible two digit endings using the digits \(1,2,3,4,5\).
Step 3: Find possible last two digits divisible by \(4\).
The two digit numbers formed from the given digits without repetition and divisible by \(4\) are
\[ 12,\;24,\;32,\;52 \] Thus, there are \(4\) possible choices for the last two digits.
Step 4: Arrange the remaining digits.
After fixing the last two digits, the remaining three digits can be arranged in
\[ 3! = 6 \] ways.
Therefore, the number of favorable five digit numbers is
\[ 4 \times 3! = 4 \times 6 = 24 \]
Step 5: Calculate the required probability.
\[ P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \] \[ P=\frac{24}{120} \] \[ P=\frac{1}{5} \]
Step 6: Final conclusion.
Hence, the required probability is
\[ \boxed{\frac{1}{5}} \]
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