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 shell may hit the target independently of the other shell 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 $\frac{x}{4}$ is.

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Detailed Solution

Let x shells are are fired at point I and (21-x) at point II. Prob (hitting the target, at least once)
$= \frac{8}{9} (1 - {(\frac{1}{2})}^{x}) + \frac{1}{9} (1 - {(\frac{1}{2})}^{21 - x}) = f (x)$ suppose
$\Rightarrow f (x) = \frac{1}{9} [9 - 8 {(\frac{1}{2})}^{x} - {(\frac{1}{2})}^{21 - x}]$

$N o w, f^{'} (x) = \frac{1}{9} [- 8 {(\frac{1}{2})}^{x} \log (\frac{1}{2}) + {(\frac{1}{2})}^{21 - x} \log (\frac{1}{2})] = \frac{1}{9} (\log 2) [{(\frac{1}{2})}^{x - 3} - {(\frac{1}{2})}^{21 - x}]$

$f^{'} (x) = 0 \Rightarrow x - 3 = 21 - x \Rightarrow x = 12 ∴ f h a s m a x . v a l u e a t x = 12 [∵ x < 12 \Rightarrow f^{'} (x) < 0 and ∵ x > 12 f^{'} (x) > 0] T h e n, \frac{x}{2} = 6$