Line 10: Line 10:
 
Second digit can be chosen in $9$ ways from $0-9$ excluding $7$ and similarly the third digit in $9$ ways.
 
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 = $8*9*9$
+
So, total no. of ways excluding $7$ = $8*9*9$
  
Total no. of ways including 7 = $9 * 10 * 10$
+
Total no. of ways including $7$ = $9 * 10 * 10$
  
 
So, ans = $(8*9*9)/(9*10*10) = 18/25$
 
So, ans = $(8*9*9)/(9*10*10) = 18/25$

Revision as of 21:20, 29 June 2014

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

(a)<math>16/25</math> (b)<math>(9/10)^3</math> (c)<math>27/75</math> (d)<math>18/25</math>

Solution by Arjun Suresh

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 $7$ = $8*9*9$

Total no. of ways including $7$ = $9 * 10 * 10$

So, ans = $(8*9*9)/(9*10*10) = 18/25$




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

(a)<math>16/25</math> (b)<math>(9/10)^3</math> (c)<math>27/75</math> (d)<math>18/25</math>

Solution by Arjun Suresh[edit]

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 $7$ = $8*9*9$

Total no. of ways including $7$ = $9 * 10 * 10$

So, ans = $(8*9*9)/(9*10*10) = 18/25$




blog comments powered by Disqus