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

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## 4 Answers

Best answer

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

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