The probability that a missile hits a target successfully is 0.75. In order to destroy the target completely, at least three successful hits are required. Then if the minimum number of missiles that have to be fired so that the probability of completely destroying the target is NOT less than 0.95, is N then find out  $\frac{N}{2}$ _____ 

We have, P = 0.75 = $\frac{3}{4}\Rightarrow q=\frac{1}{4}$
Let $P(X=r)={}^{n}c_r{\left(\frac{{\displaystyle 3}}{{\displaystyle 4}}\right)}^r{\left(\frac{{\displaystyle 1}}{{\displaystyle 4}}\right)}^{n-r}$      
We have $P(X\ge 3)\ge 0.95\Rightarrow 1-\left\{P\right(X=0)+P(X=1)+P(X=2\left)\right\}\ge 0.95$

$$\begin{gathered} 0.05\ge {}^{n}c_0{\left(\frac{1}{4}\right)}^n+{}^{n}c_1\left(\frac{3}{4}\right){\left(\frac{1}{4}\right)}^{n-1}+{}^{n}c_2{\left(\frac{3}{4}\right)}^2{\left(\frac{1}{4}\right)}^{n-2} \\ \Rightarrow \frac{1}{20}\ge \frac{1}{4^n}+\frac{3n}{4^n}+\frac{n(n-1)}{2}.\frac{9}{4^n}\Rightarrow \frac{4^n}{20}\ge 1+3n+\frac{9n^2-9n}{2} \\ \Rightarrow \frac{4^n}{20}\ge \frac{2+6n+9n^2-9n}{2}\Rightarrow \frac{4^n}{10}\ge 9n^2-3n+2\Rightarrow 2^{2n-1}\ge 5(9n^2-3n+2) \end{gathered}$$

Let  $f\left(n\right)=2^{2n-1},g\left(n\right)=5(9n^2-3n+2)$

$If n=4, f\left(4\right)=2^7=128 and$

$$\begin{gathered} g\left(4\right)=5(9{\rm X}16-3{\rm X}4+2)=670\Rightarrow n=5,f\left(5\right)=2^9=512 \\ \Rightarrow g\left(5\right)=5(9{\rm X}25-3{\rm X}5+2)=1060 \end{gathered}$$

If n=6, $f\left(6\right)=2^{11}=2048 and g\left(6\right)=5(9{\rm X}36-3{\rm X}6+2)=1540$
Hence, the minimum value of n is 6