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(a)<math>16/25</math> (b)<math>(9/10)^3</math> (c)<math>27/75</math> '''(d)<math>18/25</math>'''
 
(a)<math>16/25</math> (b)<math>(9/10)^3</math> (c)<math>27/75</math> '''(d)<math>18/25</math>'''
  
===Solution===
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==={{Template:Author|Arjun Suresh|{{arjunweb}} }}===
  
 
  First digit can be chosen in 8 ways from 1-9 excluding 7
 
  First digit can be chosen in 8 ways from 1-9 excluding 7

Revision as of 14:05, 14 April 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[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