Let A be the set of all 4-digit natural numbe | vedclass

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(C) The total number of $4$-digit numbers with exactly one $7$ is calculated as follows:If $7$ is at the thousands place: $1 	imes 9 	imes 9 	imes

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$\frac{97}{297}$

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(C) The total number of $4$-digit numbers with exactly one $7$ is calculated as follows:If $7$ is at the thousands place: $1 	imes 9 	imes 9 	imes 9 = 729$.If $7$ is not at the thousands place: $3 	imes (8 	imes 9 	imes 9) = 1944$.Total $n(S) = 729 + 1944 = 2673$.For a number to leave a remainder of $2$ when divided by $5$,the last digit must be $2$ or $7$.Case $1$: Last digit is $7$. Since exactly one $7$ is allowed,the other three digits cannot be $7$. The thousands place has $8$ choices ($1-9$ excluding $7$),and the other two places have $9$ choices each ($0-9$ excluding $7$). $n(E_1) = 8 	imes 9 	imes 9 = 648$.Case $2$: Last digit is $2$. The $7$ can be in any of the first three positions. If $7$ is at the thousands place: $1 	imes 9 	imes 9 = 81$. If $7$ is at the hundreds or tens place: $2 	imes (8 	imes 9) = 144$. $n(E_2) = 81 + 144 = 225$.Total $n(E) = 648 + 225 = 873$.Probability $P(E) = \frac{873}{2673} = \frac{97}{297}$.

Let $A$ be the set of all $4$-digit natural numbers whose exactly one digit is $7$. Then the probability that a randomly chosen element of $A$ leaves a remainder of $2$ when divided by $5$ is ..... .

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