Definition : we shall call $p$ a special prime $p_n$, if there is at least one prime of the form $2kp+1$, where $ 1 \leq k \leq n$ . it is obvious that we can call any Sophie Germain prime, a special prime $p_1$, or $p_2$, or $p_3$ ....etc. and that because we guarantee that there are a prime number of the form $2 * 1 * p +1$ .
Now, for example, if $p=29023$, and $1 \leq k \leq 20$
so, what is the probability for $p=29023$ to be a special prime $p_{20}$ ?
this is what I think, it is obvious that both of ($2*1*29023+1$, $2*2*29023+1$, $2*3*29023+1$, .......,$2*20*29023+1$) are in the interval [$2*1*29023+1, 2*20*29023+1$] ,
so, if we want to know the probability for $p=29023$ to be a special prime $p_{20}$, we should calculate the probability for a random number to be a prime in the interval [$2*1*29023+1, 2*20*29023+1$], which is $\frac{\pi [2*1*29023+1, 2*20*29023+1] }{ (2*20*29023+1)- (2*1*29023+1)}$ and then we should multiple it by $20$,
so, the probability for $p=29023$ to be a special prime $p_{20}$, is about $20 \frac{\pi [2*1*29023+1, 2*20*29023+1] }{ (2*20*29023+1)- (2*1*29023+1)}$
so, I want to know if this is the right way to calculate the probability for $p$ to be a special prime $p_n$ ? and if this is way is wrong, I need some one to show me how to calculate this probability ?
As requested in the comments, I will answer this question: given $20$ odd numbers chosen uniformly at random (and independently) from the given interval, find the probability that at least one of them is prime.
This is mostly a matter of brute force. We use a computer to determine that the number of primes in the interval is $84216$, see this.
Then it is a simple matter of determining, $p$, the probability that a (uniformly) randomly selected odd number in the interval is prime. Since there are $551437 $ odd numbers in the interval we get $$p=\frac {84216}{551437 }=.15272$$
Since the events are independent the probability that all $20$ are composite is $$(1-.15272)^{20}=.03625$$
Thus there is a greater than $96\%$ chance that at least one will be prime.
Worth remarking that a similar analysis shows that there is an $83.26\%$ chance that more than one of the will be prime. For the given list in the problem, exactly one is prime (by brute force checking) and it at least a little surprising that there is just the one.
Just to be complete, I'll mention that the number $$2\times 15\times 29023+1=870,691$$ is prime.