UC知道问两道关于概率的问题~~
来源：百度知道 编辑：UC知道 时间：2024/05/17 08:46:01
1.让M=一个年轻成年男子住在父母家中的事件 F=一个成年女子住在父母家中的事件，如果我们随意选择一个成年男子和一个成年女子，可得出
P(M)=0.56 and P(F)=0.42. 两者都住在父母家的可能性是0.24
（1）两个成年人中至少有一个选择住在父母家的概率是多少？
（2）两者都自己住（不住在父母家）的概率是多少？
原题是：letM=the event a mile young adult is living in his parents' home& F=the event a female young adult is living in her parents' home. If we randomly select a male young adult and a female young adult, the Census Bureau data enable us to conclude
P(M)=0.56 P(F)=0.42. The probability that both are living in their parents' home is 0.24.
(1)what is the probability at least one of two young adults selected is living in his or her parents' home?
(2)what is the probabilityboth young adults selected are living on their own?(neither is living in their parents' home)
2.A psychogist determined that the number of sessions required the trust of a new patient is either 1,2or3. let x be a random variable indicating the number of sessions required to gain the parents' t
P(M)=0.56 and P(F)=0.42. 两者都住在父母家的可能性是0.24
（1）两个成年人中至少有一个选择住在父母家的概率是多少？
（2）两者都自己住（不住在父母家）的概率是多少？
原题是：letM=the event a mile young adult is living in his parents' home& F=the event a female young adult is living in her parents' home. If we randomly select a male young adult and a female young adult, the Census Bureau data enable us to conclude
P(M)=0.56 P(F)=0.42. The probability that both are living in their parents' home is 0.24.
(1)what is the probability at least one of two young adults selected is living in his or her parents' home?
(2)what is the probabilityboth young adults selected are living on their own?(neither is living in their parents' home)
2.A psychogist determined that the number of sessions required the trust of a new patient is either 1,2or3. let x be a random variable indicating the number of sessions required to gain the parents' t
解（1）
整个事件为1
则两个都不和父母住的概率为：（1-0.56）*（1-0.42）=0.2552（这是根据相互独立事件的乘法分工得来的），
那么两个人至少有一个在和父母一起住的概率则为：1-0.2552=0.7448
解（2）
两个都不和父母住的概率为：（1-0.56）*（1-0.42）=0.2552（这个也是根据相互独立事件的乘法分工得来的）
一、1-（1－0.56）×（1-0.42）=0.7448
二、（1－0.56）×（1-0.42）=0.2552
p(1)=1-(1－0.56）×（1-0.42）=0.7448
p(2)=(1－0.56）×（1-0.42）=0.2552