Let A be a set of all 4 -digit natural
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:

  1. (1) $\frac{1}{5}$

  2. (2) $\frac{2}{9}$

  3. (3) $\frac{97}{297}$

  4. (4) $\frac{122}{297}$


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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