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. 0.6976
• B. 0.6776
• C. 0.5976
• D. 0.6276

Answer 1

The correct option is A

0.6976

Let $E_i$ = the event that ith shot from the guns hits the enemy i=1.2.3.4. Then according to the question.
$P\left(E_1\right)=0.4,P\left(E_2\right)=0.3,P\left(E_3\right)=0.2$ and $P\left(E_4\right)=0.1$
We have to find
$P(E_1\cup E_2\cup E_3\cup E_4)$

$P(E_1\cup E_2{E}_3\cup E_4)=1-P\left(\overline{E_1}\right)P\left(\overline{E_2}\right)P\left(\overline{E_3}\right)P\left(\overline{E_4}\right)=1-(1-0.4)(1-0.3)(1-0.2)(1-0.1)=1-0.6\times 0.7\times 0.8\times 0.9=1-0.3024=0.6976$