Question:
If a 4 digit number is formed with digits 1, 2, 3 and 5. What is the probability that the number is divisible by 25, if repetition of digits is not allowed?
If a 4 digit number is formed with digits 1, 2, 3 and 5. What is the probability that the number is divisible by 25, if repetition of digits is not allowed?
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Always check divisibility conditions for the specific divisor to fix the last digits before counting arrangements.
Updated On: Aug 27, 2025
- $\frac{1}{12}$
- $\frac{1}{4}$
- $\frac{1}{6}$
- None of these
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The Correct Option is A
Solution and Explanation
Step 1: Condition for divisibility by 25
Last two digits must be 25 or 75. From given digits {1, 2, 3, 5}, possible = 25 only (since 75 needs 7).
Step 2: Count favourable numbers
Fix last two as 25, first two from remaining {1, 3} in $2! = 2$ ways.
Step 3: Total numbers possible
Total permutations of 4 distinct digits = $4! = 24$.
Step 4: Probability
$P = \frac{2}{24} = \frac{1}{12}$.
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