Question:
Let $A$ 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
Solution:
Total cases $(4 \times 9 \times 9 \times 9)-(3 \times 9 \times 9)$
Probability $=\frac{(3 \times 9 \times 9)-(2 \times 9)+(8 \times 9 \times 9)}{\left(4 \times 9^{3}\right)-\left(3 \times 9^{2}\right)}$
$=\frac{97}{217}$
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