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  • An artillery target may be either at point I with probability or at point II with probability <

An artillery target may be either at point I with probability \frac{8}{9} or at point II with probability \frac{1}{9}. We have 21 shells each of which can be fired either at point I or II. Each shell may hit the target independently of the other shell with probability \frac{1}{2}. If the number of shells must be fired at point I to hit the target with maximum probability is \mathrm{x}, then the value of  \mathrm{12 x}, must be

Option: 1

140


Option: 2

142


Option: 3

143


Option: 4

144


Answers (1)

best_answer

Let \mathrm{A} denote the event that the target is hit when \mathrm{x} shells are fired at point I. Let \mathrm{E_1\left(E_2\right) } denote the event,

we have, \mathrm{P\left(E_1\right)=\frac{8}{9}, P\left(E_2\right)=\frac{1}{9}}

\mathrm{\Rightarrow \quad P\left(A / E_1\right)=1-\left(\frac{1}{2}\right)^x and \: P\left(A / E_2\right)=1-\left(\frac{1}{2}\right)^{21-x}}

Now, \mathrm{\quad P(A)=\frac{8}{9}\left[1-\left(\frac{1}{2}\right)^x\right]+\frac{1}{9}\left[1-\left(\frac{1}{2}\right)^{21-x}\right]}

\mathrm{ \Rightarrow \quad \frac{d P(A)}{d x}=\frac{8}{9}\left[\left(\frac{1}{2}\right)^x \log 2\right]+\frac{1}{9}\left[-\left(\frac{1}{2}\right)^{21-x} \log 2\right] }

Now, we must have \mathrm{\frac{d P(A)}{d x}=0}

\mathrm{ \Rightarrow \quad x=12 \text {, also } \frac{d^2 P(A)}{d x^2}<0 }

Hence, \mathrm{ P(A) } is maximum when \mathrm{ x=12 }

\mathrm{ \therefore \quad 12 x =12 \times 12 }

\mathrm{ =144 }

Hence option 4 is correct.



 

Posted by

Deependra Verma

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