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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 $7$
 +
 
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 $7 = 8*9*9$
+
 
Total no. of ways including $7 = 9 * 10 * 10$
+
So, total no. of ways excluding $7$ = $8*9*9$
 +
 
 +
Total no. of ways including $7$ = $9 * 10 * 10$
 +
 
 
So, ans = $(8*9*9)/(9*10*10) = 18/25$
 
So, ans = $(8*9*9)/(9*10*10) = 18/25$
  
 
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[[Category:Probability and Combinatorics]]
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[[Category: Non-GATE Questions from Probability]]
[[Category: Probability questions]]
 
[[Category:Questions]]
 

Latest revision as of 12:04, 15 July 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 = $(8*9*9)/(9*10*10) = 18/25$




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