高中数学概率题？P2=恰好成功两例的概率=3*p^2*(1-p)=0.4321C3 1*0.6*0.4*0.4=0.2882C3 2*0.6*0.6*0.4=0.432第一题：0.6*0.4*0.4*3 恰好一例成功，就是另外两例不成功 所以就那样算，乘以3是任何一例都可以。第二题：0.6*0.6*0.4*3 道理同上。那么，高中数学概率题？一起来了解一下吧。
(1)要得到60分，则必须12题全对，概率为(1/4)*(1/3)*(1/2)*(1/2)=1/48。
(2)接上，得60分的概率为1/48。
得40分的概率为(3/4)*(2/3)*(1/2)*(1/2)=1/8。
得45分的概率为(1/4)*(2/3)*(1/2)*(1/2)+(3/4)*(1/3)*(1/2)*(1/2)+(3/4)*(2/3)*(1/2)*(1/2)+(3/4)*(2/3)*(1/2)*(1/2)=17/48。
得50分的概率为(1/4)*(1/3)*(1/2)*(1/2)+(1/4)*(2/3)*(1/2)*(1/2)+(1/4)*(2/3)*(1/2)*(1/2)+(3/4)*(1/3)*(1/2)*(1/2)+(3/4)*(1/3)*(1/2)*(1/2)+(3/4)*(2/3)*(1/2)*(1/2)=17/48。
得55分的概率为(3/4)*(1/3)*(1/2)*(1/2)+(1/4)*(2/3)*(1/2)*(1/2)+(1/4)*(1/3)*(1/2)*(1/2)+(1/4)*(1/3)*(1/2)*(1/2)=7/48。
因此期望为40*(1/8)+45*(17/48)+50*(17/48)+55*(7/48)+60*(1/48)=2300/48=47又11/12。
1）要求四个题全部做对，(1/4)*(1/3)*(1/2)*(1/2) =1/48
2)
对一题概率是：(1/4)*(2/3)*(1/2)*(1/2) + (3/4)*(1/3)*(1/2)*(1/2) +2(3/4)*(2/3)*(1/2)*(1/2) = 17/48
对两题概率是：(3/4)*(2/3)*(1/2)*(1/2) + 2*(3/4)*(1/3)*(1/2)*(1/2) + 2*(1/4)*(2/3)*(1/2)*(1/2)
+ (1/4)* (1/3)*(1/2)*(1/2) = 17/48
对三题概率是：2(1/4)*(1/3)*(1/2)*(1/2) + (3/4)*(1/3)*(1/2)*(1/2) + (1/4)*(2/3)*(1/2)*(1/2) = 7/48
对三题概率是：(1/4)*(1/3)*(1/2)*(1/2) =1/48
E(x) = ( 8 + 17/48 + 2*17/48 + 3*7/48+ 4*1/48)*5 = (8+76/48)*5 =47.92
1.p=(2/5)*(3/4)*(1/3)+(3/5)*(2/4)*(1/3)=1/5
2. p=(2/5)*(2/4)*(1/3)+(2/5)*(2/4)*(1/3)+(2/5)*(2/4)*(1/3)*(1/2)+(2/5)*(1/4)*(2/3)*(1/2)+(1/5)*(2/4)*(2/3)*(1/2)+(1/5)*(2/4)*(2/3)*(1/2)+(2/5)*(2/4)*(1/3)*(1/2)+(2/5)*(1/4)*(2/3)*(1/2)=1/3
（1）1/4*1/3*1/2*1/2=1/48
（2）5*8+5*1/4+5*1/3+5*1/2+5*1/2=47.9
(1)甲以4：1获胜，意味着，一共进行了五场比赛，第五场甲胜，前四场中，甲胜3场，乙胜1场。于是：C(4,1)*0.5^4*0.5。
(2)乙获胜且比赛数大于5，意味着比赛进行了6次或7次。
6次时，乙胜4次，甲胜2次，最后一次乙胜。在前5次中，乙胜3次，甲胜2次，于是：
C(5,2)*0.5^5*0.5
7次时，乙胜4次，甲胜3次，最后一次乙胜。在前6次中，乙胜3次，甲胜3次，于是：
C(6,3)*0.5^6*0.5
(3)设比赛局数为X，则X=4,5,6,7
X=4时，p=2*0.5^4
X=5时，p=2*C(4,1)*0.5^4*0.5
X=6时，p=2*C(5,2)*0.5^5*0.5
X=7时，p=2*C(6,3)*0.5^6*0.5
之所以要*2是因为，可以甲胜也可以乙胜。
希望我的解答对你有帮助
以上就是高中数学概率题的全部内容，(1)甲以4：1获胜，意味着，一共进行了五场比赛，第五场甲胜，前四场中，甲胜3场，乙胜1场。于是：C(4,1)*0.5^4*0.5。(2)乙获胜且比赛数大于5，意味着比赛进行了6次或7次。6次时，乙胜4次，甲胜2次，最后一次乙胜。在前5次中，乙胜3次，甲胜2次，内容来源于互联网，信息真伪需自行辨别。如有侵权请联系删除。
