Target gets hit by three missiles.
The probability of hitting the target with the
– First missile is 0.4;
– Second missile is 0.5;
– Third missile is 0.7;
The target has a probability of 0.2 getting destroyed with one missile, 0.6 by two missiles and for sure (1) by three missiles.
What's the probability of the target being destroyed?
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This is the exact question on the book, teacher told us that the result is 0.458 but didn't describe the method of solving, except saying that it's not easy.
Best Answer
First: hit and destroy, Second :no hit, Third: no hit
$0.4\times 0.2 \times 0.5 \times 0.3=0.012$
First: no hit, Second :hit and destroy, Third: no hit
$0.6\times 0.5 \times 0.2 \times 0.3=0.018$
First: no hit, Second :no hit, Third: hit and destroy
$0.6\times 0.5 \times 0.7 \times 0.2=0.042$
First: no hit, (Second and Third) :hit and destroy
$0.6\times 0.5 \times 0.7 \times 0.6=0.126$
second: no hit, (first and Third) :hit and destroy
$0.5 \times 0.4 \times 0.7 \times 0.6=0.084$
third: no hit, (first and second) :hit and destroy
$0.3 \times 0.4 \times 0.5 \times 0.6=0.036$
(First, Second and Third) :hit and destroy
$0.4\times 0.5 \times 0.7 =0.14$
Add all, we get required probability as $0.458$