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Partial reply: no therapeutic
In your case the therapeutic is the wrongdoer, which makes the entire thing extremly arduous to calculate. I hope my equation for the particular case of h = 0 (no therapeutic) helps you alongside the way in which.
Minimal quantity of pictures required $n_{min}$ (all pictures hit)
$$
n_{min}= lceilfrac{x}{d}rceil
$$
Likelihood for a killing sequence of solely hits ($n_{min}$ pictures and all hit):
$$
P(n_{min}) = p^{n_{min}}
$$
Likelihood for a killing sequence with $okay$ misses ($n_{min} + okay$ pictures)
$$
p^{n_{min}}*(1-p)^{okay}
$$
Because the missed pictures can occur anyplace within the sequence, now we have to multiply this by the variety of potential permutations. Since it is a type of Binomial distribution this might be:
$$
{n_{min}+okay select n_{min}}
$$
However this accommodates additionally permutations that finish in a single or a number of missed pictures, the place the goal could be already lifeless earlier than the tip of the sequence.
So lets solely account for these permutations that finish in successful. With one occasion mounted now we have to substract 1 from high and backside.
$$
{n_{min}+k-1 select n_{min}-1}
$$
This leads us to the formular to calculate the prospect to want any quantity of pictures to kill the goal.
$$
P(n_{min}+okay) = {n_{min}+k-1 select n_{min}-1} *p^{n_{min}}*(1-p)^{okay}
$$
To get the common variety of pictures required, we multiply every likelihood with the quantity of pictures and sum every part up. Since it’s potential to overlook indefinetly, it’s an infinite sum with ever shrinking summands.
The common variety of pictures neccessary ($n_{avg}$)is :
$$
n_{avg} = sum_{okay=0}^infty( (n_{min}+okay)*P(n_{min}+okay))
$$
$$
n_{avg} = sum_{okay=0}^infty( (n_{min}+okay)*{n_{min}+k-1 select n_{min}-1} *p^{n_{min}}*(1-p)^{okay} )
$$
Perhaps another person is ready to match the therapeutic into this.
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