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 :
Select the correct option:
A
$\frac{2}{9}$
B
$\frac{122}{297}$
C
$\frac{97}{297}$
D
$\frac{1}{5}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} \mathrm{n}(\mathrm{s})= & \mathrm{n}(\text { when } 7 \text { appears on thousands place }) \\ & +\mathrm{n}(7 \text { does not appear on thousands place }) \\ = & 9 \times 9 \times 9+8 \times 9 \times 9 \times 3 \\ = & 33 \times 9 \times 9 \\ \mathrm{n}(\mathrm{E})= & \mathrm{n}(\text { last digit } 7 \text { \& } 7 \text { appears once }) \\ & +\mathrm{n}(\text { last digit } 2 \text { when } 7 \text { appears once }) \\ = & 8 \times 9 \times 9+(9 \times 9+8 \times 9 \times 2) \\ \therefore \quad \mathrm{P}(\mathrm{E})= & \frac{8 \times 9 \times 9+9 \times 25}{33 \times 9 \times 9}=\frac{97}{297}\end{aligned}$
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