GATE CSE 2017 Set 1 | Question: GA-5

Question

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. $0.9^{k}$

Comments

Here we're approximating $\frac{7}{9}*\frac{7}{10}*\frac{7}{10}*\frac{7}{10}*\frac{7}{10}....$k times

and considering it $(\frac{7}{10})^{k}$

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this question is ambiguous as none of the answer matches when we dont take 0 as left most digit..so was this question challanged and given full marking?

Answer 1

Total possibilities $=9 \times 10^{k-1},$ because most significant digit has $9$ options (excluding $0)$ and every other digit has $10$ options from $0$ to $9.$

Possibility of not containing any of the digits $0, 5, 9 = 7^{k},$ as now every digit has $7$ options.

Required probability $=\dfrac{7^{k}}{9 \times10^{k-1}} \approx  (0.7)^k$

So C is the answer.

Comments

Hello,

This is obviously incorrect because you consider 10^k total possibilities. That implies that the Most Significant digit can also have any one of 0-9 digit. But, if 0 comes in first place it would no longer be a K digit number, but, a k-1 digit number.

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Typo in above ans: Total possibilities is 9*(10^(k-1)).. Not (9^k)*(10^(k-1))

Answer 2

so C is the ans.

Answer 3

if k=1  p=.7

if k=2  p=.7 x.7 =0.7² 

in general 0.7^k ans is C

Answer 4

Probability of digits containing (0, 3 or 5) if we consider a 'k' digit number is of type base 10 = (3/10) ^k

Hence,  p(not containing 0, 3 or 5)=(7/10)^k