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NO.PZ2016062402000003 问题如下 The joint probability stribution of ranm variables X anY is given f{(x,y)}=k * x * y for x = 1,2,3, y = 1,2,3, ank is a positive constant. Whis the probability thX+Y will excee5? 1/9 1/4 1/36 Cannot termine The function x*y is scribein the following table. The sum of the entries is 36. The scaling factor k must suththe totprobability is 1. Therefore, we have k=1/36. The table shows one instanwhere X+Y 5,whiis x=3, y=3. The probability is p = 9/36 = 1/4.中文解析这道题目考察的是联合概率分布。 符合题目条件的只有一种情况(X=3,Y=3),但要想求概率,就得首先求到联合概率分布公式中的k。 题目里说,它的联合概率分布是F(x,y)=k*x*y,x可以取3种,y可以取3种,所有总共有9种可能性。其中,每一种可能性的概率都为x*y*k。将9种可能性的概率相加,它的概率之和为36k。因为所有情况的概率之和为1,所以有36k=1,所以k=1/36。 最后P(X+Y 5)= P(X=3, Y=3) = 9k =1/4 还是没懂和k的关系是什么
NO.PZ2016062402000003 问题如下 The joint probability stribution of ranm variables X anY is given f{(x,y)}=k * x * y for x = 1,2,3, y = 1,2,3, ank is a positive constant. Whis the probability thX+Y will excee5? 1/9 1/4 1/36 Cannot termine The function x*y is scribein the following table. The sum of the entries is 36. The scaling factor k must suththe totprobability is 1. Therefore, we have k=1/36. The table shows one instanwhere X+Y 5,whiis x=3, y=3. The probability is p = 9/36 = 1/4.中文解析这道题目考察的是联合概率分布。 符合题目条件的只有一种情况(X=3,Y=3),但要想求概率,就得首先求到联合概率分布公式中的k。 题目里说,它的联合概率分布是F(x,y)=k*x*y,x可以取3种,y可以取3种,所有总共有9种可能性。其中,每一种可能性的概率都为x*y*k。将9种可能性的概率相加,它的概率之和为36k。因为所有情况的概率之和为1,所以有36k=1,所以k=1/36。 最后P(X+Y 5)= P(X=3, Y=3) = 9k =1/4 为什么不是1/9?
NO.PZ2016062402000003问题如下The joint probability stribution of ranm variables X anY is given f{(x,y)}=k * x * y for x = 1,2,3, y = 1,2,3, ank is a positive constant. Whis the probability thX+Y will excee5?1/9 1/4 1/36 Cannot termine The function x*y is scribein the following table. The sum of the entries is 36. The scaling factor k must suththe totprobability is 1. Therefore, we have k=1/36. The table shows one instanwhere X+Y 5,whiis x=3, y=3. The probability is p = 9/36 = 1/4.中文解析这道题目考察的是联合概率分布。 符合题目条件的只有一种情况(X=3,Y=3),但要想求概率,就得首先求到联合概率分布公式中的k。 题目里说,它的联合概率分布是F(x,y)=k*x*y,x可以取3种,y可以取3种,所有总共有9种可能性。其中,每一种可能性的概率都为x*y*k。将9种可能性的概率相加,它的概率之和为36k。因为所有情况的概率之和为1,所以有36k=1,所以k=1/36。 最后P(X+Y 5)= P(X=3, Y=3) = 9k =1/4 解答了一年,还是不明白什么意思!什么1个K,36个K?能不能来个助教把它讲明白了
NO.PZ2016062402000003 问题如下 The joint probability stribution of ranm variables X anY is given f{(x,y)}=k * x * y for x = 1,2,3, y = 1,2,3, ank is a positive constant. Whis the probability thX+Y will excee5? 1/9 1/4 1/36 Cannot termine The function x*y is scribein the following table. The sum of the entries is 36. The scaling factor k must suththe totprobability is 1. Therefore, we have k=1/36. The table shows one instanwhere X+Y 5,whiis x=3, y=3. The probability is p = 9/36 = 1/4.中文解析这道题目考察的是联合概率分布。 符合题目条件的只有一种情况(X=3,Y=3),但要想求概率,就得首先求到联合概率分布公式中的k。 题目里说,它的联合概率分布是F(x,y)=k*x*y,x可以取3种,y可以取3种,所有总共有9种可能性。其中,每一种可能性的概率都为x*y*k。将9种可能性的概率相加,它的概率之和为36k。因为所有情况的概率之和为1,所以有36k=1,所以k=1/36。 最后P(X+Y 5)= P(X=3, Y=3) = 9k =1/4 反复看了解答,还是没明白为什么不是1/3*1/3=1/9
NO.PZ2016062402000003 问题如下 The joint probability stribution of ranm variables X anY is given f{(x,y)}=k * x * y for x = 1,2,3, y = 1,2,3, ank is a positive constant. Whis the probability thX+Y will excee5? 1/9 1/4 1/36 Cannot termine The function x*y is scribein the following table. The sum of the entries is 36. The scaling factor k must suththe totprobability is 1. Therefore, we have k=1/36. The table shows one instanwhere X+Y 5,whiis x=3, y=3. The probability is p = 9/36 = 1/4.中文解析这道题目考察的是联合概率分布。 符合题目条件的只有一种情况(X=3,Y=3),但要想求概率,就得首先求到联合概率分布公式中的k。 题目里说,它的联合概率分布是F(x,y)=k*x*y,x可以取3种,y可以取3种,所有总共有9种可能性。其中,每一种可能性的概率都为x*y*k。将9种可能性的概率相加,它的概率之和为36k。因为所有情况的概率之和为1,所以有36k=1,所以k=1/36。 最后P(X+Y 5)= P(X=3, Y=3) = 9k =1/4 不懂为什么是1/36,函数不是k*x*y?x有3种可能性;y有3种可能性,不应该是1/9吗?