我有一個大學的機率數學題 ><!急
when calculating all the values of a binomial distribution the work can usually be simplified by first calculating b (0;n, θ) and then using the recursion formula b(x+1;n,θ)=θ(n-x)/(x-1)(1-θ)*b(x;n,θ)
verify this formula and use it to calculate the values of the binomial distribution with n=7 and θ=0.25
please steps by steps !
Update:我打錯一個地方，以下是我的修正。
b(x+1;n,θ)=θ(n-x)/(x+1)(1-θ)*b(x;n,θ)
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b(x+1;n,θ)=θ(n-x)/(x-1)(1-θ)*b(x;n,θ) 錯了，應該是
b(x+1;n,θ)=θ(n-x)/【(x+1)(1-θ)】*b(x;n,θ)
2011-07-15 08:19:48 補充：
b(x+1;n,θ)=C(n,x+1) θ^(x+1)*(1-θ)^(n-x-1)b(x;n,θ)=C(n,x) θ^(x)*(1-θ)^(n-x)b(x+1;n,θ)/ b(x;n,θ)= C(n,x+1)/ C(n,x)* θ/(1-θ)=(n-x)/(x+1) * θ/(1-θ)b(x+1;n,θ)=(θ(n-x))/((x+1)(1-θ))*b(x;n,θ)
n=7, θ=0.25,1-θ=0.75, (θ(n-x))/((x+1)(1-θ))=(0.25/0.75)*((7-x)/(x+1))b (0,0;7, 0.25)=0.25^0*0.75^7=0.75^7=0.1335x=0, b (1,0;7, 0.25)= (0.25/0.75)*((7-0)/(0+1))* 0.75^7=7*0.25*0.75^6= C(7,1)* 0.25*0.75^6=0.3115x=1, b (2,0;7, 0.25)= (0.25/0.75)*((7-1)/(1+1))* 7*0.25*0.75^6=7*6*0.25^2*0.75^5/(1*2)=C(7,2)* 0.25^2*0.75^5=0.3115x=2, b (3,0;7, 0.25)= (0.25/0.75)*((7-2)/(2+1))* 7*6*0.25^2*0.75^5/2=7*6*5*0.25^3*0.75^4/(1*2*3)=C(7,3)* 0.25^3*0.75^4=0.1730x=3, b (4,0;7, 0.25)= (0.25/0.75)*((7-3)/(3+1))* 7*6*5*0.25^3*0.75^4/(2*3)=7*6*5*4*0.25^4*0.75^3/(1*2*3*4)=C(7,4)* 0.25^4*0.75^3=0.0577x=4, b (5,0;7, 0.25)= (0.25/0.75)*((7-4)/(4+1))* 7*6*5*4*0.25^4*0.75^3/(2*3*4)=7*6*5*4*3*0.25^5*0.75^2/(1*2*3*4*5)=C(7,5)* 0.25^5*0.75^2=0.0115x=5, b (6,0;7, 0.25)= (0.25/0.75)*((7-5)/(5+1))* 7*6*5*4*3*0.25^5*0.75^2/(2*3*4*5)=7*6*5*4*3*2*0.25^6*0.75^1/(1*2*3*4*5*6)=C(7,6)* 0.25^6*0.75^1=0.0013x=6, b (7,0;7, 0.25)= (0.25/0.75)*((7-6)/(6+1))* 7*6*5*4*3*2*0.25^6*0.75^1/(2*3*4*5*6)=7*6*5*4*3*2*1*0.25^7*0.75^0/(1*2*3*4*5*6*7)=C(7,7)* 0.25^7*0.75^0=0.0001