An artillery target may be either at point `I` with probability `8/9` or at point `II` with probability `1/9.` We have `55` shells, each of which can be fired either rat point `I` or `II.` Each shell may hit the target, independent of the other shells, with probability `1/2.` Maximum number of shells must be fired at point `I` to have maximum probability is

Options

A. 20
B. 25
C. 29
D. 35

Answer

C

Explanation

Let A denote the event that target is hig when x shells are fired at point I. Let `P_(1) and P_(2)` denote the event that the target is at point I and II, respectively. We have `P(P_(1))=8//9,P(P_(2))=1//9,P(A//P_(1))=1-(1//2)^(x),P(A//P_(2))=1-(1//2)^(55-x).`
Now from total probability theorem,
`P(A)=P(P_(1))P(A//P)+P(P_(2))=1-(1//2)`
`=1/9(8-8((1)/(2))^(x)+1-((1)/(2))^(55-x))`
`=1/9(9-8((1)/(2))^(x)-((1)/(2))^(55-x))`
Now, `(dP(A))/(dx)=1/9(-8((1)/(2))^(x)ln((1)/(2))+((1)/(2))^(55-x)ln((1)/(2)))`
(Note that in this step, it is being assumed that `x in R^(+)`)
`=1/9ln((1)/(2))((1)/(2))^(55-x)(1-((1)/(2))^(2x-58))`
`gt0if xlt29`
`lt0if xgt29`
Therefore, P(A) is maximum at x = 29, Thus, 29 sells must be friend at point I.