概率问题袋中有红球、白球共100
设袋中有x个红球，则取出3个球全为同色的概率：
p=(x/100)((x-1)/99)((x-2)/98)+((100-x)/100)((99-x)/99)((98-x)/98)
=(x(x-1)(x-2)+(100-x)(99-x)(98-x))/(100*99*98)
=((x-1)((x-1)^2-1)+(99-x)((99-x)^2-1))/(100*99*98)
=((x-1)^3-x+1+(99-x)^3-99+x))/(100*99*98)
=(x^3-3x^2+3x-1+99^3-3*99^2x+3*99x^2-x^3-98)/(100*99*98)
=(3*98x^2-3...全部
设袋中有x个红球，则取出3个球全为同色的概率：
p=(x/100)((x-1)/99)((x-2)/98)+((100-x)/100)((99-x)/99)((98-x)/98)
=(x(x-1)(x-2)+(100-x)(99-x)(98-x))/(100*99*98)
=((x-1)((x-1)^2-1)+(99-x)((99-x)^2-1))/(100*99*98)
=((x-1)^3-x+1+(99-x)^3-99+x))/(100*99*98)
=(x^3-3x^2+3x-1+99^3-3*99^2x+3*99x^2-x^3-98)/(100*99*98)
=(3*98x^2-3*98*100x+100*99*98))/(100*99*98)
=(x^2-100x+3300)/3300
=((x-50)^2+800)/3300
故当x=50时，p最小，pmin=800/3300=8/33
即当袋中有50个红球时，取出的3个球全为同色的概率最小。
收起
