袋中有2008顆球,分別編號1,2,3,......,2008,每顆球被取出的機率相等,任取3顆球,這3顆球中編號最大者的編號為x,求x的期望值?謝謝! 從某處看到的蠻有意思的題目,答案是6027/4,但是我不會。
2008颗球任取三颗,共 C 2008 3 {\displaystyle C_{2008}^{3}} 种取法; 三颗中最大值为1和2不可能,从最大者为3开始至2008,编号最大为 x {\displaystyle x} 时,有 C x − 1 2 {\displaystyle C_{x-1}^{2}} 种取法; 期望值: ∑ x = 3 2008 x ⋅ C x − 1 2 C 2008 3 {\displaystyle \sum _{x=3}^{2008}x\cdot {\frac {C_{x-1}^{2}}{C_{2008}^{3}}}} 计算中会用到: C b a = b ! a ! ⋅ ( b − a ) ! {\displaystyle C_{b}^{a}={\frac {b!}{a!\cdot (b-a)!}}} ∑ x = 1 n x 2 = n ( n + 1 ) ( 2 n + 1 ) / 6 {\displaystyle \sum _{x=1}^{n}x^{2}=n(n+1)(2n+1)/6} ∑ x = a b x = ( a + b ) ( b − a + 1 ) / 2 {\displaystyle \sum _{x=a}^{b}x=(a+b)(b-a+1)/2} 数字比较大但应该还是可以手算的。
是否還要用到 ( b + 1 a + 1 ) = ( b a ) + ( b a + 1 ) {\displaystyle {b+1 \choose a+1}={b \choose a}+{b \choose a+1}} ?
( 2008 3 ) {\displaystyle {2008 \choose 3}} 是常数,可以直接拆出来。 x ( x − 1 2 ) = x ! 2 ! ( x − 3 ) ! = 1 2 x ( x − 1 ) ( x − 2 ) = 1 2 ( x 3 − 3 x 2 + 2 x ) {\displaystyle x{{x-1} \choose 2}={\frac {x!}{2!(x-3)!}}={\frac {1}{2}}x(x-1)(x-2)={\frac {1}{2}}(x^{3}-3x^{2}+2x)} 。常数拆出来,每一项单独求和。
∑ x = 3 2008 x ⋅ ( x − 1 2 ) ( 2008 3 ) = ∑ x = 3 2008 ( x ⋅ ( x − 1 ) ! 2 ! ⋅ ( x − 3 ) ! ⋅ 6 2008 ⋅ 2007 ⋅ 2006 ) = ∑ x = 3 2008 ( x ! 3 ! ⋅ ( x − 3 ) ! ⋅ 6 ⋅ 3 2008 ⋅ 2007 ⋅ 2006 ) = 18 2008 ⋅ 2007 ⋅ 2006 ⋅ ∑ x = 3 2008 ( x 3 ) {\displaystyle \sum _{x=3}^{2008}x\cdot {\frac {x-1 \choose 2}{2008 \choose 3}}=\sum _{x=3}^{2008}\left(x\cdot {\frac {(x-1)!}{2!\cdot (x-3)!}}\cdot {\frac {6}{2008\cdot 2007\cdot 2006}}\right)=\sum _{x=3}^{2008}\left({\frac {x!}{3!\cdot (x-3)!}}\cdot {\frac {6\cdot 3}{2008\cdot 2007\cdot 2006}}\right)={\frac {18}{2008\cdot 2007\cdot 2006}}\cdot \sum _{x=3}^{2008}{x \choose 3}} 單獨算 ∑ x = 3 2008 ( x 3 ) {\displaystyle \sum _{x=3}^{2008}{x \choose 3}} = ( 3 3 ) + ( 4 3 ) + ( 5 3 ) + ( 6 3 ) + ⋯ + ( 2008 3 ) {\displaystyle ={3 \choose 3}+{4 \choose 3}+{5 \choose 3}+{6 \choose 3}+\cdots +{2008 \choose 3}} = ( 4 4 ) + ( 4 3 ) + ( 5 3 ) + ( 6 3 ) + ⋯ + ( 2008 3 ) {\displaystyle ={4 \choose 4}+{4 \choose 3}+{5 \choose 3}+{6 \choose 3}+\cdots +{2008 \choose 3}} = ( 5 4 ) + ( 5 3 ) + ( 6 3 ) + ⋯ + ( 2008 3 ) {\displaystyle ={5 \choose 4}+{5 \choose 3}+{6 \choose 3}+\cdots +{2008 \choose 3}} = ( 6 4 ) + ( 6 3 ) + ⋯ + ( 2008 3 ) {\displaystyle ={6 \choose 4}+{6 \choose 3}+\cdots +{2008 \choose 3}} = ( 7 4 ) + ⋯ + ( 2008 3 ) {\displaystyle ={7 \choose 4}+\cdots +{2008 \choose 3}} = ( 2008 4 ) + ( 2008 3 ) {\displaystyle ={2008 \choose 4}+{2008 \choose 3}} = ( 2009 4 ) {\displaystyle ={2009 \choose 4}} 期望值 = 18 2008 ⋅ 2007 ⋅ 2006 ⋅ ( 2009 4 ) = 18 2008 ⋅ 2007 ⋅ 2006 ⋅ 2009 ⋅ 2008 ⋅ 2007 ⋅ 2006 24 = 6027 4 {\displaystyle ={\frac {18}{2008\cdot 2007\cdot 2006}}\cdot {2009 \choose 4}={\frac {18}{2008\cdot 2007\cdot 2006}}\cdot {\frac {2009\cdot 2008\cdot 2007\cdot 2006}{24}}={\frac {6027}{4}}}