Let A denote the event that a 6-digit integer | vedclass

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(B) The total number of $6$-digit integers formed using the digits ${0, 1, 2, 3, 4, 5, 6}$ without repetition is given by choosing the first digit in

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$\frac{4}{9}$

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(B) The total number of $6$-digit integers formed using the digits ${0, 1, 2, 3, 4, 5, 6}$ without repetition is given by choosing the first digit in $6$ ways (excluding $0$) and the remaining $5$ digits in $P(6, 5)$ ways.$n(S) = 6 	imes 6! = 4320$.For a number to be divisible by $3$,the sum of its digits must be divisible by $3$. The sum of all digits ${0, 1, 2, 3, 4, 5, 6}$ is $21$. To form a $6$-digit number,we exclude one digit such that the sum of the remaining $6$ digits is divisible by $3$. The excluded digit must be $0, 3,$ or $6$.Case $I$: Exclude $0$. The digits are ${1, 2, 3, 4, 5, 6}$. Number of ways $= 6! = 720$.Case $II$: Exclude $3$. The digits are ${0, 1, 2, 4, 5, 6}$. The first digit cannot be $0$,so there are $5 	imes 5! = 600$ ways.Case $III$: Exclude $6$. The digits are ${0, 1, 2, 3, 4, 5}$. The first digit cannot be $0$,so there are $5 	imes 5! = 600$ ways.Total favourable cases $n(A) = 720 + 600 + 600 = 1920$.Probability $P(A) = \frac{n(A)}{n(S)} = \frac{1920}{4320} = \frac{192}{432} = \frac{4}{9}$.

Let $A$ denote the event that a $6$-digit integer formed by $0, 1, 2, 3, 4, 5, 6$ without repetitions is divisible by $3$. Then the probability of event $A$ is equal to:

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