题目
以X记某医院一天出生的婴儿的个数,Y记其中男婴的个数,设X和Y-|||-的联合分布律为-|||- X=n,Y=m =dfrac ({e)^-14((7.14))^m((6.86))^m-m}(m!(n-m)!)-|||-=0,1,2,... ,n ; =0, 1,2,····-|||-(1)求边缘分布律.-|||-(2)求条件分布律.-|||-(3)特别,写出当 X=20 时,Y的条件分布律.
题目解答
答案
解析
步骤 1：求边缘分布律
为了求出边缘分布律，我们需要分别计算 $P\{ X=n\}$ 和 $P\{ Y=m\}$。根据联合分布律，我们有：
$P\{ X=n\} =\sum _{m=0}^{n}P\{ X=n,Y=m\} $
$P\{ Y=m\} =\sum _{n=m}^{\infty }P\{ X=n,Y=m\} $
步骤 2：计算 $P\{ X=n\}$
$P\{ X=n\} =\sum _{m=0}^{n}\dfrac {{e}^{-14}{(7.14)}^{m}{(6.86)}^{n-m}}{m!(n-m)!}$
$=\dfrac {{e}^{-14}}{n!}{(7.14+6.86)}^{n}=\dfrac {14^{n}{e}^{-14}}{n!}$ ,n=0 ,1,2,...
步骤 3：计算 $P\{ Y=m\}$
$P\{ Y=m\} =\sum _{n=m}^{\infty }\dfrac {{e}^{-14}{(7.14)}^{m}{(6.86)}^{n-m}}{m!(n-m)!}$
$=\dfrac {{e}^{-14}}{m!}{(7.14)}^{m}{e}^{6.86}=\dfrac {{(7.14)}^{m}{e}^{-7.14}}{m!}$ , $m=0,1,2,\cdots $
步骤 4：求条件分布律
对于 m=0 ,1,2 ,.,
$P\{ X=n|Y=m\} =\dfrac {P\{ X=n,Y=m\} }{P\{ Y=m\} }$
$=\dfrac {{e}^{-14}{(7.14)}^{m}{(6.86)}^{n-m}}{m!(n-m)!}/\dfrac {{(7.14)}^{m}}{m!}{e}^{-7.14}$
$=\dfrac {{(6.86)}^{n-m}}{(n-m)!}{e}^{-6.86}$ , $n=m,m+1,\cdots $
对于 n=0 ,1,2 ,.,,
$P\{ Y=m|X=n\} =\dfrac {P\{ X=n,Y=m\} }{P\{ X=n\} }$
$=\dfrac {{e}^{-14}{(7.14)}^{m}{(6.86)}^{n-m}}{m!(n-m)!}/\dfrac {{14}^{n}{e}^{-14}}{n!}$
$=\dfrac {n!}{m!(n-m)!}{(\dfrac {7.14}{14})}^{m}{(\dfrac {6.86}{14})}^{n-m}$
$=\dfrac {n!}{m!(n-m)!}{(0.51)}^{m}{(0.49)}^{n-m}$ , $m=0,1,\cdots ,n$
步骤 5：当 X=20 时,Y的条件分布律
$P\{ Y=m|X=20\} =\dfrac {20!}{m!(20-m)!}{(0.51)}^{m}{(0.49)}^{20-m}$ , m=0,1 ,.,20.
为了求出边缘分布律，我们需要分别计算 $P\{ X=n\}$ 和 $P\{ Y=m\}$。根据联合分布律，我们有：
$P\{ X=n\} =\sum _{m=0}^{n}P\{ X=n,Y=m\} $
$P\{ Y=m\} =\sum _{n=m}^{\infty }P\{ X=n,Y=m\} $
步骤 2：计算 $P\{ X=n\}$
$P\{ X=n\} =\sum _{m=0}^{n}\dfrac {{e}^{-14}{(7.14)}^{m}{(6.86)}^{n-m}}{m!(n-m)!}$
$=\dfrac {{e}^{-14}}{n!}{(7.14+6.86)}^{n}=\dfrac {14^{n}{e}^{-14}}{n!}$ ,n=0 ,1,2,...
步骤 3：计算 $P\{ Y=m\}$
$P\{ Y=m\} =\sum _{n=m}^{\infty }\dfrac {{e}^{-14}{(7.14)}^{m}{(6.86)}^{n-m}}{m!(n-m)!}$
$=\dfrac {{e}^{-14}}{m!}{(7.14)}^{m}{e}^{6.86}=\dfrac {{(7.14)}^{m}{e}^{-7.14}}{m!}$ , $m=0,1,2,\cdots $
步骤 4：求条件分布律
对于 m=0 ,1,2 ,.,
$P\{ X=n|Y=m\} =\dfrac {P\{ X=n,Y=m\} }{P\{ Y=m\} }$
$=\dfrac {{e}^{-14}{(7.14)}^{m}{(6.86)}^{n-m}}{m!(n-m)!}/\dfrac {{(7.14)}^{m}}{m!}{e}^{-7.14}$
$=\dfrac {{(6.86)}^{n-m}}{(n-m)!}{e}^{-6.86}$ , $n=m,m+1,\cdots $
对于 n=0 ,1,2 ,.,,
$P\{ Y=m|X=n\} =\dfrac {P\{ X=n,Y=m\} }{P\{ X=n\} }$
$=\dfrac {{e}^{-14}{(7.14)}^{m}{(6.86)}^{n-m}}{m!(n-m)!}/\dfrac {{14}^{n}{e}^{-14}}{n!}$
$=\dfrac {n!}{m!(n-m)!}{(\dfrac {7.14}{14})}^{m}{(\dfrac {6.86}{14})}^{n-m}$
$=\dfrac {n!}{m!(n-m)!}{(0.51)}^{m}{(0.49)}^{n-m}$ , $m=0,1,\cdots ,n$
步骤 5：当 X=20 时,Y的条件分布律
$P\{ Y=m|X=20\} =\dfrac {20!}{m!(20-m)!}{(0.51)}^{m}{(0.49)}^{20-m}$ , m=0,1 ,.,20.