Find the probability that a number selected at random between 100 and 999 (both inclusive) will not contain the digit 7.

The probability that a number selected at random between100 and 999 (both inclusive) will not contain the digit 7 is 

(A) $\frac{16}{9}$, 

(B) $\left(\frac{9}{10}\right)^3$, 

(C) $\frac{27}{75}$, 

(D) $\frac{18}{25}$. 

(GATE 1995)

Answer: (D) $\frac{18}{25}$. 

Explanation: 

All the numbers between 100 and 999 (both inclusive) have three digits. 

No. of ways of choosing the left-most digit from 1 to 9 excluding 7 = 8. 

No. of ways of choosing the middle digit from 0 to 9 excluding 7 = 9. 

No. of ways of choosing the right-most digit from 0 to 9 excluding 7 = 9. 

So, the total number of ways of selecting a number between 100 and 999 (both inclusive) that will not contain the digit 7 is = $8 \times 9 \times 9$. 

Again, the total number of ways of selecting any number between 100 and 999 (both inclusive) is = $^{900}C_1 = 900$. 

Therefore, the required probability is = $\frac{8 \times 9 \times 9}{900} = \frac{18}{25}$.