if n>=2m, 数学期望值=2m if n<2m, 数学期望值= n |
此题不那么像狐月月以前所说的 "小儿科" 。建议狐月月先温习一下什么叫数学期望。 |
For any boy, Let P be the probability that he is next to a girl, then the answer to the problem is N*P. P is: 2/(M+N)* M/(M+N-1) + (M+N-2)/(M+N)*(1-(N-1)*(N-2)/(M+N-1)*(M+N-2)) |
按概率(机会大小)平均。 |
结果正确。但表达式不规范(on the priority of the operations),漏了一些括号,不易被理解,甚至会被理解成错误结果(我相信HF自己有正确的理解,要得到这样一个只需添点括号就正确的表达式,必有正确的思维过程)。正确结果的规范表达式可以是: P=2m/(m+n)(m+n-1)+(m+n-2)[1-(n-1)(n-2)/(m+n-1)(m+n-2)]/(m+n)。 然后,用 "随机变量和的数学期望等于它们数学期望的和" 这一定理,得到所求数学期望为nP。 |
完全正确! |
For any boy, Let P be the probability that he is next to at least one girl, then the answer to the problem is N*P. 1. If N=1, E=1; 2. if N >1: For any one boy the possiblity that he is next to no girls Q=2(N-1)(M+N-2)!/(M+N)!+(N-1) 3. Therefore, E=N*P=N*(1-Q). 4. You can verify the solution by assuming that N=2, E=2M(M+3)/(M+1)/(M+2). |
For any boy, Let P be the probability that he is next to at least one girl, then the answer to the problem is E=N*P. 1. two simple cases If N=1, E=1; If N=2, E=2M(M+3)/[(M+1)(M+2)] 2. if N >2: For any one boy the possiblity that he is next to no girls Q=2(N-1)(M+N-2)!/(M+N)!+(N-1)(N-2)(M+N-2)!/(M+N)! =N(N-1)/[M+N)(M+N-1)] 3. Therefore, E=N*P=N*(1-Q). 4. The solution in 3 also applies to N=1 and 2. 本贴由[数学期望]最后编辑于:2007-2-26 12:55:14 本贴由[数学期望]最后编辑于:2007-2-26 12:56:10 本贴由[数学期望]最后编辑于:2007-2-26 13:1:37 |
当m和n趋近无穷大(或是无穷大)时, 数学期望还有定义(或意义)吗?(我在瞎想,呵呵) |
有。有限但趋向无穷时取极限就行。无穷时以概率分布代替用古典概型求出来的那些概率来计算数学期望。 |
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