设随机变量X～b(3,0.4),且随机变量Y=X（3-X）/2，则p{X=Y}= 设随机变量X~B(3,0.4),且随机变量Y=(3-X)/2...

\u8bbe\u968f\u673a\u53d8\u91cfx~b(3,0.2),\u4e14\u968f\u673a\u53d8\u91cfY=X(3-X)/2,\u5219P(Y=0)=

P(Y=0)=P(X=3)+P(X=0)=0.2^3+0.8^3=0.52\u6ee1\u610f\u8bf7\u597d\u8bc4\u54e6\uff0c\u6211\u6982\u7387\u8bba\u8003\u5b8c\u4e86\uff0c\u563b\u563b\uff0c\u795d\u4f60\u8003\u8bd5\u987a\u5229\u5566

E(X)=3\u00d70.4=1.2
E(Y)=3/2-1/2\u00b7E(X)=1.5-0.6=0.9
\u3010\u9644\u6ce8\u3011
\uff081\uff09X~B(n\uff0cp)\uff0c\u5219 E(X)=np
\uff082\uff09E(kX+b)=kE(X)+b

找出能满足条件的X取值即可求出概率，过程如图。经济数学团队帮你解答，请及时采纳。谢谢！

璁鹃殢鏈哄彉閲廥锝濨(3,0.4)
绛旓細浜岄」鍒嗗竷锛孍x=3*0.4=1.2锛

璁鹃殢鏈哄彉閲廥锝瀊(3,0.4),涓旈殢鏈哄彉閲廦=X(3-X)/2,鍒檖{X=Y}=
绛旓細鎵惧嚭鑳芥弧瓒虫潯浠剁殑X鍙栧煎嵆鍙眰鍑烘鐜囷紝杩囩▼濡傚浘銆傜粡娴庢暟瀛﹀洟闃熷府浣犺В绛旓紝璇峰強鏃堕噰绾炽傝阿璋紒

璁鹃殢鏈哄彉閲廥~B(3, 0.4),, 鍒橮(Y=1)鐨勫间负
绛旓細X~B(3,0.4)锛屽垯P(X=1)=C(3,1)(0.4)(0.6^2)=0.432 P(X鈮1)=1-P(X=0)=1-C(2 0)*(p^0)(1-p)^2=1-(1-p)^2=3/4 p=1/2 p(y鈮1)=1-P(Y=0)=1-C(3 0)*(p^0)(1-p)^3=7/8

璁鹃殢鏈哄彉閲廥~B(3,0.4),涓旈殢鏈哄彉閲廦=(3-X)/2,鍒橢(Y)=
绛旓細E(X)=3脳0.4=1.2 E(Y)=3/2-1/2路E(X)=1.5-0.6=0.9 銆愰檮娉ㄣ戯紙1锛塜~B(n,p),鍒 E(X)=np 锛2锛塃(kX+b)=kE(X)+b

闅忔満鍙橀噺X~b(3,0.4)琛ㄧず浠涔 ? 杩欎釜寮忓瓙琛ㄧず浠涔堟剰鎬濆晩 ~~姹傛暀~~_鐧 ...
绛旓細X鏈嶄粠鍙傛暟涓3,0.4鐨凚ernoulli鍒嗗竷 鍗冲湪3娆＄嫭绔嬮噸澶嶈瘯楠屼腑,姣忔璇曢獙鎴愬姛鐨勬鐜囦负0.4,鎬荤殑鎴愬姛娆℃暟涓X鐨鍒嗗竷 P(X=k)=C(3,k)*0.4^k*0.6^(3-k)

璁鹃殢鏈哄彉閲廥~B(3,0.4),涓旈殢鏈哄彉閲廦=(3-X)/2,鍒橢(Y)=
绛旓細E(X)=3脳0.4=1.2 E(Y)=3/2-1/2路E(X)=1.5-0.6=0.9 銆愰檮娉ㄣ戯紙1锛塜~B(n锛宲)锛屽垯 E(X)=np 锛2锛塃(kX+b)=kE(X)+b

璁鹃殢鏈哄彉閲弜~b(3,0.4)鍒檖(x鈮1)
绛旓細浣犲ソ锛乆鏄簩椤瑰垎甯冿紝鎵浠(X鈮1)=1-P(X<1)=1-P(X=0)=1-0.6^3=0.784銆傜粡娴庢暟瀛﹀洟闃熷府浣犺В绛旓紝璇峰強鏃堕噰绾炽傝阿璋紒

璁鹃殢鏈哄彉閲廥~B(3,0.4),姹侾{X鈮1}
绛旓細X灞炰簬n锛3,p锛0.4鐨勪簩椤瑰垎甯冿紝鎵浠{X锛0}锛0.6^3锛0.216 P{X鈮1}锛1-P{X锛0}锛0.784 鍥炵瓟瀹屾瘯姹傞噰绾

璁綳~B(3,0.4),姹備笅鍒闅忔満鍙橀噺鐨勬暟瀛︽湡鏈
绛旓細x鏈嶄粠b(3,0.4)锛屾晠x鍙彇鍊间负0锛1,2,3 褰搙=0鏃讹紝y=0 褰搙=1,y=-1 褰搙=2锛寉=0 褰搙=3锛寉=3 鎵浠ワ紝y鏄釜绂绘暎鍨闅忔満鍙橀噺锛屽彲鍙栫殑鍊间负-1,0,3 p锛坹=-1锛=p(x=1)=c(3,1)(0.3)^1*(0.7)^2=0.441 p锛坹=0锛=p锛坸=0)+p锛坸=2锛=0.343+0.189=0.532 p锛坹=...

姒傜巼璁X~B(3,0.4)浠涔堟剰鎬
绛旓細X鈫B鎰忔濇槸闅忔満鍙橀噺X 鏈嶄粠浜岄」鍒嗗竷銆锛3 0.4锛鎰忔濇槸鍋氫笁娆＄嫭绔嬮噸澶嶅疄楠岋紝浜嬩欢鍙戠敓鐨勬鐜囦负0.4銆傚鏋滄眰鏁板鏈熸湜灏辨槸3涔0.4

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