Let A be a set of all 4-digit natural numbers whose exactly one digit is 7.

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 :

Accepted Answer

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)

$$\therefore$$ $$P(E) = {{8 \times 9 \times 9 + 9 \times 25} \over {33 \times 9 \times 9}} = {{97} \over {297}}$$

Solutions · 1