GATE CS 2017 Set 1 Q.60 (GA Q.60): Quantitative Aptitude

The probability that a $k$ -digit number does NOT contain the digits 0, 5, or 9 is

$0.3^k$

$0.6^k$

$0.7^k$

$0.9^k$

Official answer and solution

Solution

Calculate the probability that an arbitrary digit position does not contain 0, 5, or 9, and apply the product rule across all k independent digits.

In a $k$ -digit sequence where each digit is chosen uniformly from the 10 decimal digits $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ :

The total number of possible digits for each position is 10.

The digits that must NOT appear are $\{0, 5, 9\}$ , which is a set of 3 digits.

Therefore, the number of permissible digits for any given position is:

10 - 3 = 7 \quad (\text{namely } \{1, 2, 3, 4, 6, 7, 8\})

The probability that a single digit is permissible is:

P(\text{valid digit}) = \frac{7}{10} = 0.7

Since the choice of each of the $k$ digits is independent, the probability that none of the $k$ digits are 0, 5, or 9 is:

P = (0.7)^k = 0.7^k

Final answer

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