# Answer in Statistics and Probability for Iqra #90338

Let vector r = (x1, x2, x3) ( where xi = 1 if missile i hit the target, and xi = 0 otherwise) represent outcome of firing.

All outcomes favorable for event A = at least two missiles hit the target are:

(1, 1, 1), probability P(1 ,1, 1) = 0.4*0.5*0.6 = P(0, 0, 0)

(1, 1, 0), probability P(1 ,1, 0) = 0.4*0.5*(1-0.6) = P(1, 0, 0)

(1, 0, 1), probability P(1 ,0, 1) = 0.4*(1-0.5)*0.6 = P(0, 1, 0)

(0, 1, 1), probability P(0 ,1, 1) = (1-0.4)*0.5*0.6 = P(0, 0, 1)

Sum of probabilities in the left column gives probability P(A)

Sum of probabilities in the right column gives probability P(A’) (where A’ is complementary event)

P(A) = P(A’) and P(A) + P(A’) =1 implies 2*P(A) = 1 and P(A) = 1/2

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It is possible to get answer simpler, reformulate problem :

Three missiles are fired at a target. The probabilities of _missing_ target are (1 – 0.4) = 0.6, (1 – 0.5) = 0.5 and (1 – 0.6) = 0.4 respectively . If missiles are fired independently . What is the probability that at least 2 missiles _miss_ the target?

And we have problem with the same probabilities. Events in questions of original and reformulated problem complementary,

then sum of their probabilities is 1, and since probabilities of these events are equal answer is 1/2.

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Answer: Probability that at least two missile hit the target is 1/2.

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