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 :
Solution
Total cases :<br><br>$\underline 6$ . $\underline 6$ . $\underline 5$ . $\underline 4$ . $\underline 3$ . $\underline 2$<br><br>n(s) = 6 . 6!<br><br>Favourable cases :<br><br>Number divisible by 3 $\equiv$ Sum of digits must be divisible by 3<br><br>Case - I<br><br>1, 2, 3, 4, 5, 6<br><br>Number of ways = 6!<br><br>Case - II<br><br>0, 1, 2, 4, 5, 6<br><br>Number of ways = 5 . 5!<br><br>Case - III<br><br>0, 1, 2, 3, 4, 5<br><br>Number of ways = 5 . 5!<br><br>n(favourable) = 6! + 2 . 5 . 5!<br><br>$P = {{6! + 2.\,5.\,5!} \over {6\,.\,6!}} = {4 \over 9}$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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