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

Answer is (D)

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

Comments

nice explanation @Arjun sir

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 ...

Answer 4

We have 3 places to fill, first find all the three digits number containing 7.
Fill 7 at only one place:
7 _ _: we've 2 places to fill, without mixing up more 7s (i.e using these 9 digits 0,1,2,3,4,5,6,8,9), possible numbers are: 9 * 9 = 81
_ 7 _: we've 2 places to fill but here the first digit can't be 0, so 8 digits for first place and 9 digits for the last place excluding 7, possible numbers are: 8 * 9 = 72.
_ _ 7: using the same explanation as above, possible numbers are: 8 * 9 = 72.

Fill 7 at 2 places:
7 7 _: 9 digits possible for the last place excluding 7, so possible numbers = 9
7 _ 7: 9 digits possible for the middle place excluding 7, so possible numbers = 9
_ 7 7: First digit can't be 0, so 8 digits possible excluding 7, so possible numbers = 8

Fill 7 at all 3 places:
7 7 7: possible numbers = 1

Therefore, total numbers containing 7 = 81 + 72 + 72 + 9 + 9 + 8 + 1 = 252
Total numbers between 100 and 999 = 900
Therefore, total numbers not containing 7 = 900 -252 = 648
Therefore, probability that a number selected at random will not contain 7 = 648 / 900 = 18 /25