A 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 shall may hit the target independently of the other shall with probability $\frac{1}{2}$ . The number of shells must be fired at point I to hit the target with maximum probability is x, then $\frac{x}{2}$ is _____

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answer is 6.

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Detailed Solution

Let x shells are fired at point I and (21–x) at point II.
Prob (hitting the target, at least once )
$\begin{array}{l} = \frac{8}{9} (1 - {(\frac{1}{2})}^{x}) + \frac{1}{9} (1 - {(\frac{1}{2})}^{21 - x}) = f (x) suppose \\ \Rightarrow f (x) = \frac{1}{9} [9 - 8 {(\frac{1}{2})}^{x} - {(\frac{1}{2})}^{21 - x}] \end{array}$
$\begin{array}{l} Now f^{1} (x) = \frac{1}{9} [- 8 {(\frac{1}{2})}^{x} \log (\frac{1}{2}) + {(\frac{1}{2})}^{21 - x} \log (\frac{1}{2})] \\ = \frac{1}{9} (\log 2) [{(\frac{1}{2})}^{x - 3} - {(\frac{1}{2})}^{21 - x}] \\ f^{1} (x) = 0 \Rightarrow x - 3 = 21 - x \Rightarrow x = 12 \\ ∴ f has max. Value at x = 12 \\ [∵ x < 12 \Rightarrow f^{1} (x) < 0 and \\ ∵ x > 12 \Rightarrow f^{1} (x) > 0] \end{array}$
$Then \frac{x}{2} = 6$