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

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

                                                                          

Answer 2

0.7^k

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

Answer 3

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  

Answer 4

Total K positions, and to make it a K-digit number the 1^st  position will not have 0 (so 9 choices to fill 1^st position i.e. 1 to 9) while the remaining K-1 positions can be filled in 10 ways (0 to 9). So, total K-digit numbers that we can have = 9*10^(k-1).

#cases in which number does NOT contain the digits 0,5 or 9 = 7

so, probability = 7^k/9*10^(k-1) = 1.11(0.7)^k