搜搜考试网条件概率的问题设苹果树上开N朵花是随机事件,且服从概率分布函数P[N=n]=(1-p)p^n,p属于(0,1)区间,又假
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条件概率的问题
设苹果树上开N朵花是随机事件,且服从概率分布函数P[N=n]=(1-p)p^n,p属于(0,1)区间,又假设每朵花结果的概率的是a,且各朵花之间结苹果是独立的.
现在树上有r个苹果,请问树上原先有n朵花的概率是多少?
回复nowusing：
我开始的思路基本和你一样，在你的这一步中“P(R=r) = C(N',r)(1-ap)^(N'-r)(ap)^r”，
你是假设N‘是已知的，所以依照你的思路可以解题，但是这是未知的。还有，开花的概率你理解的有一点问题
我再给你描述下英文的原题吧：
The number of flowers N appearing on an apple tree is a random variable,with distribution P[N=n]=(1-p)p^n,for some p属于(0,1).Assume that a flower turns into a fruit with probability a independently from the other flowers on the tree.
Given that tree there is a number r of apples on the tree,what is the probability that originally there were n flowers on it?
设苹果树上开N朵花是随机事件,且服从概率分布函数P[N=n]=(1-p)p^n,p属于(0,1)区间,又假设每朵花结果的概率的是a,且各朵花之间结苹果是独立的.
现在树上有r个苹果,请问树上原先有n朵花的概率是多少?
回复nowusing：
我开始的思路基本和你一样，在你的这一步中“P(R=r) = C(N',r)(1-ap)^(N'-r)(ap)^r”，
你是假设N‘是已知的，所以依照你的思路可以解题，但是这是未知的。还有，开花的概率你理解的有一点问题
我再给你描述下英文的原题吧：
The number of flowers N appearing on an apple tree is a random variable,with distribution P[N=n]=(1-p)p^n,for some p属于(0,1).Assume that a flower turns into a fruit with probability a independently from the other flowers on the tree.
Given that tree there is a number r of apples on the tree,what is the probability that originally there were n flowers on it?
![条件概率的问题设苹果树上开N朵花是随机事件,且服从概率分布函数P[N=n]=(1-p)p^n,p属于(0,1)区间,又假](/uploads/image/z/4437574-70-4.jpg?t=%E6%9D%A1%E4%BB%B6%E6%A6%82%E7%8E%87%E7%9A%84%E9%97%AE%E9%A2%98%E8%AE%BE%E8%8B%B9%E6%9E%9C%E6%A0%91%E4%B8%8A%E5%BC%80N%E6%9C%B5%E8%8A%B1%E6%98%AF%E9%9A%8F%E6%9C%BA%E4%BA%8B%E4%BB%B6%2C%E4%B8%94%E6%9C%8D%E4%BB%8E%E6%A6%82%E7%8E%87%E5%88%86%E5%B8%83%E5%87%BD%E6%95%B0P%5BN%3Dn%5D%3D%281-p%29p%5En%2Cp%E5%B1%9E%E4%BA%8E%280%2C1%29%E5%8C%BA%E9%97%B4%2C%E5%8F%88%E5%81%87)
okay.I think I got it this time.
The probability distribution actually makes sense since
P(N=0)+P(N=1)+...=(1-p)(1+p+p^2+p^3+...)=1
N:number of flowers
R:number of apples
By Bayesian rule,P(N=n|R=r) = P(R=r|N=n)*P(N=n) / P(R=r) (1)
P[N=n]= (1-p)p^n (2)
P(R=r|N=n)=C(n,r)a^r (1-a)^(n-r) (3) -there is no unknowns in this equation
Now we only need to solve P(R=r).
Notice that P(R=r) = P(R=r|N=n)*P(N=n) + P(R=r|Nn)*P(Nn) (4) - first term is also known by (2)&(3)
And P(Nn)=1-P(N=n) = 1 - (1-p)p^n (5)
Now we only need to solve P(R=r|Nn).Notice N>=r and n>=r.
P(R=r|Nn) = P(R=r|N=r)+P(R=r|N=r+1)+...+P(R=r|N=n-1)+P(R=r|N=n+1)+ ...
= -P(R=r|N=n)+P(R=r|N=r)+P(R=r|N=r+1)+...
=-C(n,r)a^r (1-a)^(n-r) + [P(R=r|N=r)+P(R=r|N=r+1)+...] (6)
Now we only need to solve P(R=r|N=r)+P(R=r|N=r+1)+...
P(R=r|N=r)+P(R=r|N=r+1)+...=a^r+C(r+1,r)a^r(1-a)+.C(r+2,r)a^r(1-a)^2
=a^r[1+C(r+1,r)(1-a)+C(r+2,r)(1-a)^2+...]
=a^r * [1+(1-a)]^r (7) - using Taylor's expansion to get this
Finally,plug(2)-(7) to (1) would solve the problem.
Hope this helps!
The probability distribution actually makes sense since
P(N=0)+P(N=1)+...=(1-p)(1+p+p^2+p^3+...)=1
N:number of flowers
R:number of apples
By Bayesian rule,P(N=n|R=r) = P(R=r|N=n)*P(N=n) / P(R=r) (1)
P[N=n]= (1-p)p^n (2)
P(R=r|N=n)=C(n,r)a^r (1-a)^(n-r) (3) -there is no unknowns in this equation
Now we only need to solve P(R=r).
Notice that P(R=r) = P(R=r|N=n)*P(N=n) + P(R=r|Nn)*P(Nn) (4) - first term is also known by (2)&(3)
And P(Nn)=1-P(N=n) = 1 - (1-p)p^n (5)
Now we only need to solve P(R=r|Nn).Notice N>=r and n>=r.
P(R=r|Nn) = P(R=r|N=r)+P(R=r|N=r+1)+...+P(R=r|N=n-1)+P(R=r|N=n+1)+ ...
= -P(R=r|N=n)+P(R=r|N=r)+P(R=r|N=r+1)+...
=-C(n,r)a^r (1-a)^(n-r) + [P(R=r|N=r)+P(R=r|N=r+1)+...] (6)
Now we only need to solve P(R=r|N=r)+P(R=r|N=r+1)+...
P(R=r|N=r)+P(R=r|N=r+1)+...=a^r+C(r+1,r)a^r(1-a)+.C(r+2,r)a^r(1-a)^2
=a^r[1+C(r+1,r)(1-a)+C(r+2,r)(1-a)^2+...]
=a^r * [1+(1-a)]^r (7) - using Taylor's expansion to get this
Finally,plug(2)-(7) to (1) would solve the problem.
Hope this helps!
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