Let $$S$$ be the sample space of all five digit numbers. If $$p$$ is the probability that a randomly selected number from $$S$$, is a multiple of $$7$$ but not divisible by $$5$$, then $$9p$$ is equal to

Let

$$S=\{\text{all five digit numbers}\}$$

Total number of five digit numbers is

$$99999-10000+1=90000$$

Hence,

$$n(S)=90000$$

Now count five digit numbers divisible by $$7.$$

Smallest five digit multiple of $$7$$ is

$$10003$$

Largest five digit multiple of $$7$$ is

$$99995$$

Using

$$n=\frac{l-a}{d}+1,$$

number of multiples of $$7$$ is

$$n=\frac{99995-10003}{7}+1$$

$$=\frac{89992}{7}+1$$

$$=12856+1$$

$$=12857$$

Now count numbers divisible by both $$7$$ and $$5,$$ i.e., divisible by $$35.$$

Smallest five digit multiple of $$35$$ is

$$10010$$

Largest five digit multiple of $$35$$ is

$$99995$$

Hence,

$$n=\frac{99995-10010}{35}+1$$

$$=\frac{89985}{35}+1$$

$$=2571+1$$

$$=2572$$

Therefore, favorable numbers are

$$12857-2572=10285$$

Thus,

$$p=\frac{10285}{90000}$$

$$=\frac{2057}{18000}$$

Hence,

$$9p=\frac{2057}{2000}$$

Therefore,the required value is

$$\boxed{1.0285}$$.