Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Join our Q&A platform to connect with experts dedicated to providing precise answers to your questions in different areas. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
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
We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
