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? Choose the correct choice.
A) 0.6976	B) 0.7013	C) 0.789	D) 0.8356	E) 0.8666
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.   0.6976lternate 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>A</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?