Let $\mathrm{A}$ be a set of all 4-digit natural numbers whose exactly one digit is 7 . Then the probability that a randomly chosen element of $A$ leaves remainder 2 when divided by 5 is:
Correct Option: , 3
$\mathrm{n}(\mathrm{s})=\mathrm{n}$ (when 7 appears on thousands place)
$+\mathrm{n}(7$ does not appear on thousands place)
$=9 \times 9 \times 9+8 \times 9 \times 9 \times 3$
$=33 \times 9 \times 9$
$\mathrm{n}(\mathrm{E})=\mathrm{n}$ (last digit $7 \& 7$ appears once)
$+\mathrm{n}$ (last digit 2 when 7 appears once)
$=8 \times 9 \times 9+(9 \times 9+8 \times 9 \times 2)$
$\therefore P(E)=\frac{8 \times 9 \times 9+9 \times 25}{33 \times 9 \times 9}=\frac{97}{297}$
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