GATE1995-1.18

Question

The probability that a number selected at random between $100$ and $999$ (both inclusive) will not contain the digit $7$ is:

(a)$\dfrac{16}{25}$

(b)$\left(\dfrac{9}{10}\right)^{3}$

(c)$\dfrac{27}{75}$

(d)$\dfrac{18}{25}$

Comments

can someone show how to calculate it in other way i.e. (1-p(contains 7))?

Answer 1

First digit can be chosen in $8$ ways from $1−9$ excluding $7$

Second digit can be chosen in $9$ ways from $0−9$ excluding $7$ and similarly the third digit in $9$ ways.

So, total no. of ways excluding $7 = 8\times 9\times 9$

Total no. of ways including $7 = 9\times 10\times 10$

So, ans $= \dfrac{(8\times 9\times 9)}{(9\times 10\times10)}=\dfrac{18}{25}$

Answer 2

8*9*9/9*10*10=18/25 as we can use 8 more digits in hundreds place,9 more digit in tens place and 9 digits in units place

Comments

simple!!

Answer 3

Number of 3 digit numbers with atleast one 7

          = (10*10) + (10*9) + (10*9) - 10 - 10 - 9 + 1

         = 100 + 90 + 90 - 10 - 10 - 9 + 1

         = 252

Number of 3 digit numbers with no 7 in any of its digits

        = 900 - 252

        = 648

Probability(Number not having digit 7)

        = 648 / 900

        = 18 / 25.

Option D) is the answer ...