JEE MAIN - Mathematics (2021 - 25th February Evening Shift - No. 5)
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 :
$${2 \over 9}$$
$${1 \over 5}$$
$${122 \over 297}$$
$${97 \over 297}$$
Explanation
n(s) = n(when 7 appears on thousands place)
+ 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
n(E) = n(last digit 7 & 7 appears once)
+n(last digit 2 when 7 appears once)
= 8 $$\times$$ 9 $$\times$$ 9 + (3 $$\times$$ 9 $$\times$$ 9 - 2 $$\times$$ 9)
$$\therefore$$ $$P(E) = {{8 \times 9 \times 9 + 9 \times 25} \over {33 \times 9 \times 9}} = {{97} \over {297}}$$
+ 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
n(E) = n(last digit 7 & 7 appears once)
+n(last digit 2 when 7 appears once)
= 8 $$\times$$ 9 $$\times$$ 9 + (3 $$\times$$ 9 $$\times$$ 9 - 2 $$\times$$ 9)
$$\therefore$$ $$P(E) = {{8 \times 9 \times 9 + 9 \times 25} \over {33 \times 9 \times 9}} = {{97} \over {297}}$$
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