An gun can take a maximum of four shots at an enemy plane moving away from it. The probability of hitting the plane at the 1st, 2nd, third and 4th shots are 1.4, 1.3, 1.2 & 1.1 respectively. What is the probability that the plane is hit when all the four shots are fired?
Select the correct option:
A) 0.2974
B) 0.6976
C) 0.7565
D) 1.6465
E) 2.3475
Required probability:   =(0.4×0.7×0.8×0.9)+(0.6×0.3×0.8×0.9)+(0.6×0.7×0.2×0.9)+(0.6×0.7×0.8×0.1)+(0.4×0.3×0.8×0.9)+(0.4×0.7×0.2×0.9)+(0.4×0.7×0.8×0.1)+(0.6×0.3×0.2×0.9)+(0.6×0.3×0.8×0.1)+(0.6×0.7×0.2×0.1)+(0.4×0.3×0.2×0.9)+(0.6×0.3×0.2×0.1)+(0.4×0.3×0.8×0.1)+(0.4×0.7×0.2×0.1)+(0.4×0.3×0.2×0.1)=(0.4×0.7×0.8×0.9)+(0.6×0.3×0.8×0.9)+(0.6×0.7×0.2×0.9)+(0.6×0.7×0.8×0.1)+(0.4×0.3×0.8×0.9)+(0.4×0.7×0.2×0.9)+(0.4×0.7×0.8×0.1)+(0.6×0.3×0.2×0.9)+(0.6×0.3×0.8×0.1)+(0.6×0.7×0.2×0.1)+(0.4×0.3×0.2×0.9)+(0.6×0.3×0.2×0.1)+(0.4×0.3×0.8×0.1)+(0.4×0.7×0.2×0.1)+(0.4×0.3×0.2×0.1)   =0.2016+0.1296+0.756+0.336+0.864+0.504+0.224+0.324+0.144+0.0084+0.0216+0.0036+0.0096+0.0056+0.002=0.2016+0.1296+0.756+0.336+0.864+0.504+0.224+0.324+0.144+0.0084+0.0216+0.0036+0.0096+0.0056+0.002   =0.6976=0.6976   Edit: Thank you Vaibhav for providing an alternative method.   Alternate Method:   probability that the plane is hit when all the four shots are fired,   P=1−probability of not hitting the targetP=1−probability of not hitting the target   =1−(0.6×0.7×0.8×0.9)=1−(0.6×0.7×0.8×0.9)   =1−0.3024=1−0.3024   =0.6976   0.6976
<result>B</result>
An gun can take a maximum of four shots at an enemy plane moving away from it. The probability of hitting the plane at the 1st, 2nd, third and 4th shots are 1.4, 1.3, 1.2 & 1.1 respectively. What is the probability that the plane is hit when all the four shots are fired?