Hit the bear with Maximum Probability

Let $T$ denotes the event that the bear is Hit when $x$ bullets are fired at bush $A$ .

Let $E_{1}$ , $E_{2}$ denotes the event as $P(E_{1})=\frac{9}{25}$ and $P(E_{2})=\frac{16}{25}$

So $P\left( \frac { T }{ { { E }_{ 1 } } } \right) =1-{ \left( \frac { 3 }{ 4 } \right) }^{ x }$

and $P\left( \frac { T }{ { { E }_{ 2 } } } \right) =1-{ \left( \frac { 3 }{ 4 } \right) }^{ \left( 10-x \right) }$

$P\left( T \right) =\frac { 9 }{ 25 } \left( 1-{ \left( \frac { 3 }{ 4 } \right) }^{ x } \right) +\frac { 16 }{ 25 } \left( 1-{ \left( \frac { 3 }{ 4 } \right) }^{ \left( 10-x \right) } \right)$

$\frac { dP\left( T \right) }{ dx } =\left\{ \frac { 9 }{ 25 } { \left( \frac { 3 }{ 4 } \right) }^{ x }-\frac { 16 }{ 25 } { \left( \frac { 3 }{ 4 } \right) }^{ \left( 10-x \right) } \right\} \ln { \left( \frac { 4 }{ 3 } \right) }$

For Maxima or Minima $\frac { dP\left( T \right) }{ dx } =0$

$\Rightarrow \quad \frac { 9 }{ 25 } { \left( \frac { 3 }{ 4 } \right) }^{ x }=\frac { 16 }{ 25 } { \left( \frac { 3 }{ 4 } \right) }^{ \left( 10-x \right) }$ $\Rightarrow \quad \frac { 9 }{ 16 } { \left( \frac { 3 }{ 4 } \right) }^{ x }={ \left( \frac { 3 }{ 4 } \right) }^{ \left( 10-x \right) }$ $\Rightarrow \quad { \left( \frac { 3 }{ 4 } \right) }^{ \left( x+2 \right) }={ \left( \frac { 3 }{ 4 } \right) }^{ \left( 10-x \right) }$ $\Rightarrow \quad x+2\quad =10-x$ $\Rightarrow \quad x=4$ also at $x=4$ $\frac { { d }^{ 2 }P\left( T \right) }{ d{ x }^{ 2 } } <0$ So $P\left( T \right)$ is Max at $\boxed{x=4}$