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) 0.3^k
(B) 0.6^k
(C) 0.7^k
(D) 0.9^k

Answer: (C)
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
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