Question

An anti-aircraft gun can take a maximum of four shots at an enemy plane moving away from it.The probabilities of hitting the plane at the first, second, third and fourth shot are $0.4,0.3,0.2$ and $0.1$ respectively. What is the probability that the gun hits the plane?

• A. $0.9976$
• B. $0.6976$
• C. $0.0024$
• D. $0.3024$

Answer 1

The correct option is B $0.6976$

Let $E_1$ be the event that the first shot from the gun hits the enemy .
$E_2$ be the event that the second shot from the gun hits the enemy
$E_3$ be the event that the third shot from the gun hits the enemy
$E_4$ be the event that the fourth shot from the gun hits the enemy
Given $P\left(E_1\right)=0.4$
$\Rightarrow P\left({\overline{E}}_1\right)=0.6$
$P\left(E_2\right)=0.3$
$\Rightarrow P\left({\overline{E}}_2\right)=0.7$
$P\left(E_3\right)=0.2$
$\Rightarrow P\left({\overline{E}}_3\right)=0.8$
and $P\left(E_4\right)=\mathrm{0.1.}$
$\Rightarrow P\left({\overline{E}}_4\right)=0.9$
We have to find $P(E_1\cup E_2\cup E_3\cup E_4).$
$=1-P\left(\overline{E_1\cup E_2\cup E_3\cup E_4}\right)$
$=1-P(\overline{E_1}\cap \overline{E_2}\cap \overline{E_3}\cap \overline{E_4})$
$=1-P\left(\overline{E_1}\right)P\left(\overline{E_2}\right)P\left(\overline{E_3}\right)P\left(\overline{E_4}\right).$ (Since,$E_1,E_2,E_3,E_4$ are independent, so their complements also.)
$=1-0.6\times 0.7\times 0.8\times 0.9$
$=1-0.3024$
$P(E_1\cup E_2\cup E_3\cup E_4)=0.6976$