二项分布

题目

二项分布
P{X=k}=(5)0.1^k0.9^5-k,
k
请问当k=0,1,2,3,4,5时,相对应的概率是多少,主要是详细的计算过程,谢谢!
第二行的k在第一个5的下面

答案

主要是算出（m,n）m在上,n在下
（m,n）＝m!/[n!*(m-n)!]
所以
P{X=0}=(5,0)*0.1^0*0.9^5=[5!/(0!*(5-0)!)]*0.1^0*0.9^5=0.9^5=0.59049
P{X=1}=(5,1)*0.1^1*0.9^4=[5!/(1!*(5-1)!)]*0.1^1*0.9^4=5*0.1^1*0.9^4=0.32805
P{X=2}=(5,2)*0.1^2*0.9^3=[5!/(2!*(5-2)!)]*0.1^2*0.9^3=10*0.1^2*0.9^3=0.0729
P{X=3}=(5,3)*0.1^3*0.9^2=[5!/(3!*(5-3)!)]*0.1^3*0.9^2=10*0.1^3*0.9^2=0.0081
P{X=4}=(5,4)*0.1^4*0.9^1=[5!/(4!*(5-4)!)]*0.1^4*0.9^1=5*0.1^4*0.9^1=0.00045
P{X=5}=(5,5)*0.1^5*0.9^0=[5!/(5!*(5-5)!)]*0.1^5*0.9^0=0.00001