{"id":47,"date":"2020-01-29T20:04:08","date_gmt":"2020-01-29T20:04:08","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=47"},"modified":"2020-10-13T16:21:37","modified_gmt":"2020-10-13T16:21:37","slug":"crazy-jumping-frog","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=47","title":{"rendered":"Crazy Jumping Frog"},"content":{"rendered":"\n\"Crazy\n\n\n\n
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Crazy Jumping Frog can jump forward at a distance either 1 foot or 2 feet. The probability it jumps 1 foot forward is $p$, and the probability it jumps 2 feet is $1-p$. Assume that each jump is independent and the frog has infinite amount of energy \ud83d\ude42<\/p>\n

    \n
  1. What the expected distance after the frog making $n$ jumps?<\/li>\n
  2. If there is an uncovered well $n$ feet away from the frog, what is the probability that the frog fells into the well if it jumps forward trying to pass the well?<\/li>\n<\/ol>\n

    Click here for the solutions.<\/a><\/p>\n

    \n

    Solution for Question 1<\/strong> The expected distance in feet for making $1$ jump is $$1\\cdot p + 2\\cdot (1-p)=2-p$$ Since each jump is independent, therefore, the expect distance after the frog making $n$ jumps is $n(2-p)$ feet.<\/p>\n

    Solution for Question 2<\/strong> Denote the probability of the frog jumping $n$ feet is $f(n)$, we have $$f(n)=p\\cdot f(n-1) + (1-p)\\cdot f(n-2)\\tag{1}$$ with initial values as $$f(0)=1$$ $$f(1)=p$$ $$f(2)=p\\cdot f(1) + (1-p)\\cdot f(0)=p^2-p+1$$
    \nTo solve (1), which is a     Recurrence Relation<\/a>, we need to find the roots of its characteristic equation<\/a> as the following: $$x^2=px+(1-p)$$\nSolve the above equation, we have\n$$x_{1}=1, x_{2}=p-1$$\nTherefore\n$$f(n)=c_{1}x_{1}^n+c_{2}x_{2}^n=c_{1}+c_{2}(p-1)^n$$\nBecause $f(0)=1$, $f(1)=p$, we have $$\n\\begin{array}{lcc@}\nc_{1} + c_{2} & = & 1\\\\\nc_{1} + c_{2}\\cdot (p-1) & = & p\n\\end{array}\n$$\nSolve the above linear equations, we have\n$$c_{1}=\\frac{1}{2-p}, \\ \\ \\ \\ c_{2}=\\frac{1-p}{2-p}$$\nTherefore\n$$f(n)=c_{1}+c_{2}(p-1)^n=\\frac{1}{2-p}+\\frac{1-p}{2-p}(p-1)^n=\\frac{1-(p-1)^{n+1}}{2-p}$$\n<\/p>\n<\/div>\n\n\n\n

    <\/div><\/div>\n\n\n\n

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