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Sagot :
Solution :
The probability of winning when you choose n is = [tex]$^nC_2\left(\frac{1}{2}\right)^n$[/tex]
[tex]$n\left(\frac{n-1}{2}\right)\times \left(\frac{1}{2}\right)^n = n(n-1)\left(\frac{1}{2}\right)^{n+1}$[/tex]
Apply log on both the sides,
[tex]$f(n) = \log\left((n)(n-1)\left(\frac{1}{2}\right)^{n-1}\right) = \log n +\log (n-1)+(n+1) \ \log\left(\frac{1}{2}\right)$[/tex]
Differentiation, f(x) is [tex]$f'=\frac{1}{x}+\frac{1}{(x-1)}+\log\left(\frac{1}{2}\right)$[/tex]
Let us find x for which f' is positive and x for which f' is negative.
[tex]$\frac{1}{x}+\frac{1}{(x-1)} > 0.693$[/tex] , since [tex]$\log(1/2) = 0.693147$[/tex]
For x ≤ 3, f' > 0 for [tex]$\frac{1}{x}+\frac{1}{x-1}+\log\left(\frac{1}{2}\right)>0$[/tex]
[tex]$\frac{1}{x}+\frac{1}{x-1}-0.6931470$[/tex]
That means f(x) is increasing function for n ≤ 3
[tex]$\frac{1}{x}+\frac{1}{x-1}< 0.693147 $[/tex] for x > 4
f' < 0 for n ≥ 4, that means f(n) is decreasing function for n ≥ 4.
Probability of winning when you chose n = 3 is [tex]$3(3-1)\left(\frac{1}{2}\right)^{3+1}=0.375$[/tex]
Probability of winning when you chose n = 4 is [tex]$4(4-1)\left(\frac{1}{2}\right)^{4+1}=0.375$[/tex]
Therefore, we should chose either 3 or 4 to maximize chances of winning.
The probability of winning with an optimal choice is n = 0.375
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