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 :
JEE · Math · previous-year question
- A.$${4 \over {9}}$$correct
- B.$${9 \over {56}}$$
- C.$${11 \over {27}}$$
- D.$${3 \over {7}}$$
Answer
A. $${4 \over {9}}$$
Explanation
Total cases : $$\underline 6 $$ . $$\underline 6 $$ . $$\underline 5 $$ . $$\underline 4 $$ . $$\underline 3 $$ . $$\underline 2 $$ n(s) = 6 . 6! Favourable cases : Number divisible by 3 $$ \equiv $$ Sum of digits must be divisible by 3 Case - I 1, 2, 3, 4, 5, 6 Number of ways = 6! Case - II 0, 1, 2, 4, 5, 6 Number of ways = 5 . 5! Case - III 0, 1, 2, 3, 4, 5 Number of ways = 5 . 5! n(favourable) = 6! + 2 . 5 . 5! $$P = {{6! + 2.\,5.\,5!} \over {6\,.\,6!}} = {4 \over 9}$$
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