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An artillery target may be either at point A with probability 8/9 or at point B withprobability 1/9. We have 21 shells each of which can be fixed either at point A or B. Each shell may hit the target independently of the other shell with probability 1/2 . How many shells must be fired at point A to hit the target with maximum probabilty ?

Debjit Dutta , 13 Years ago
Grade 12
anser 1 Answers
SHAIK AASIF AHAMED

Last Activity: 10 Years ago

Hello student,
Please find the answer to your question below
Let A denote the event that the target is hit when x shells are fired at point I
We have p(e1)=8/9 and p(e2)=1/9
p(a/e1)=1-(1/2)xand p(a/e2)=1-(1/2)21-x
Now p(a)=8/9(1-(1/2)x)+1/9(1-(1/2)21-x)
By differentiating the above equation and making it equal to zero we get x=12
So 12 shells must be fired at point A to hit the target with maximum probabilty

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