Question:
Let A denote the event that a 6-digit integer formed by 0, 1, 2, 3, 4, 5, 6 without repetitions, be divisible by 3. Then probability of event A is equal to :
Let A denote the event that a 6-digit integer formed by 0, 1, 2, 3, 4, 5, 6 without repetitions, be divisible by 3. Then probability of event A is equal to :
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A 6-digit number from these 7 digits omits exactly one digit; divisibility by 3 depends on which digit is omitted.
Updated On: Nov 4, 2026
- \(\frac{4}{9}\)
- \(\frac{9}{56}\)
- \(\frac{11}{27}\)
- \(\frac{3}{7}\)
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The Correct Option is A
Solution and Explanation
Sum of the digits \(0+1+\dots+6 = 21\). A 6-digit number uses all digits except one, say \(d\); its digit sum is \(21 - d\), divisible by 3 iff \(d\in\{0,3,6\}\).
Total numbers: omit 0: \(6! = 720\); omit any of 1..6 (6 choices): first digit non-zero gives \(6! - 5! = 600\) each. Total \(= 720 + 3600 = 4320\).
Favourable: omit 0: 720; omit 3: 600; omit 6: 600. Total \(= 1920\).
\(P(A) = \frac{1920}{4320} = \frac{4}{9}\).
Answer: (1) \(\frac{4}{9}\)
Total numbers: omit 0: \(6! = 720\); omit any of 1..6 (6 choices): first digit non-zero gives \(6! - 5! = 600\) each. Total \(= 720 + 3600 = 4320\).
Favourable: omit 0: 720; omit 3: 600; omit 6: 600. Total \(= 1920\).
\(P(A) = \frac{1920}{4320} = \frac{4}{9}\).
Answer: (1) \(\frac{4}{9}\)
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