# GATE | GATE-CS-2017 (Set 1) | Question 4

The probability that a k-digit number does NOT contain the digits 0, 5 or 9 is
(A) 0.3k
(B) 0.6k
(C) 0.7k
(D) 0.9k

Explanation:
For 10 digits(0 to 9) total possible cases = (10)k
Excluding digits 0, 5, 9, total possible cases = (7)k with numbers 1,2,3,4,6,7,8
Required probability = (7)k/(10)k= (0.7)k
Therefore option C is correct

Alternate Solution

Case(1): When 0 can not be as left most digit:
Then sample space = 9 * 10^{k-1}
Favorable event = 7^{k}
Required probability = ( 7^{k} ) / ( 9 * 10^{k-1}).
None option matches.

Case(2): When 0 can be as left most digit:
Then sample space = 10^{k}
Favorable event = 7^{k}
Required probability = ( 7^{k} ) / ( 10^{k} ) = (7/10)^{k} = (0.7)^{k}.
Option (c) matches.

This solution is contributed by Mithlesh Upadhyay

Quiz of this Question

My Personal Notes arrow_drop_up
Article Tags :
Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
0 Average Difficulty : 0/5.0