Q80.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 : (1) 9 (2) 4 56 9 (3) 3 (4) 11 7 27
What This Question Tests
This problem combines permutation principles with the divisibility rule for 3 to calculate the probability of forming a 6-digit number, without repetition, that is divisible by 3.
Concepts Tested
Formulas Used
P(A) = n(A)/n(S)
Divisibility rule for 3: sum of digits is divisible by 3
Number of permutations P(n,k) = n!/(n-k)!
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📋 Question Details
- Chapter
- Probability
- Topic
- Probability of an event with permutations
- Year
- 2021
- Shift
- 16 Mar Shift 2
- Q Number
- Q80
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 13: Probability | Class 11 Mathematics Ch 7: Permutation & Combination
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