Probability problem involving binomial formula.

Question

Every shooting independently from all the others hit the ship with probability of $p = 0,2.$ With $k$ hits the ship sinks with probability $1-p_2^k,$ where $0 < p_2 < 1.$ Calculate the probability that the ship will sink if it will be hit on n times, where $n=5, p_2=0,5$. My solution: if $p=0,2$, then $q=0,8.$ $$P_5(0 \le k \le n)=\left(\frac{4}{5}\right)^5+5\cdot \left( \frac{4}{5}\right)^4\cdot\frac{1}{5}+10 \cdot \left(\frac{4}{5}\right)^3 \cdot \frac{1}{5^2}+10 \cdot \left(\frac{4}{5}\right)^2 \cdot \frac{1}{5^3}+5\cdot \left(\frac{4}{5}\right)\cdot \frac{1}{5^4}=\frac{3124}{3125} \approx 0,999.$$ I know my solution is incorrect, because I don't even use $p_2$ and the probability is too high.

Answer 1

There is a little problem. You are not taking into account that if the ship is to be destroyed in, let's say, the second shot it couldn't have been destroyed in the first one. Also I am no quite sure why you are using the binomial distribution.... maybe if you extend you explain your train of thought a little bit.

Anyhow, the prbability is then $$ P=\sum_{k=1}^5P_k $$ Where $P_k=P(\text{sank in } k)=(1-p^k)P(\text{not sank in }k-1)=(1-p^k)(1-\sum_{l=1}^{k-1}P(\text{sank in } l))$. Then $$ P_1=1-p,\\ P_2=(1-p^2)(1-(1-p))=(1-p^2)p\\ P_3=(1-p^3)(1-(1-p)-(1-p^2)p)=(1-p^3)p^3\\ P_4=(1-p^4)(1-(1-p)-(1-p^2)p-(1-p^3)p^3)=(1-p^4)p^6\\ P_5=\dots=(1-p^5)p^{10} $$

Adding all up I got 0.999969.