# GATE CSE 2017 Set 1 | Question: GA-5

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

1. $0.3^{k}$
2. $0.6^{k}$
3. $0.7^{k}$
4. $0.9^{k}$

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

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

Total possibilities =$(10)^{k},$ because every digit has $10$ options from $0\;to\; 9.$

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

Asked probability =$\dfrac{(7)^{k}}{(10)^{k}} = (0.7)^k$

1 flag:
✌ Edit necessary (prithatiti “if 0 comes in first place it would no longer be a K digit number, but, a (k-1) digit number.”)

edited
14
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. so C is the ans.

if k=1  p=.7

if k=2  p=.7 x.7 =0.7

in general 0.7k ans is C

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

let p be probability that an ith place in a number  contains digits 0,3 or 5.

then p=3/10

~p = 7/10

therefore, probability that a k digit number doesn't contain digits 0,3 or 5 = (~p)(~p)................k times = (7/10)k

0.7^k

When digit is one digit number we will have 7 possibility.

Probablity=   required outcomes total outcomes

No of ways to form k-digits with no 0, 5, 9 (with repetitions allowed) = 7^k

Total no of ways to for k-digit number( with repetitions allowed)= 9.(10^k-1)

hence probablity=   7^k / 9.(10^k-1)  == 1.11(0.7^k)  similar to option C

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