The stimulus mentioned the numbers 1 to 6. That's a smog. The only thing matters is odd or even among the numbers. Since "At least three of them have rolled an even number", let's do it one step at a time. If all 6 dices are even, there is only one way to do it (all six gamblers get even numbers) and there is one way in which the three specific gamblers have all even. If 5 dices are even, there are 6 ways to do it (only one gambler gets an odd number) and there are 3 ways in which the three gamblers have all even. If 4 dices are even, there are 15 ways to do it and there are 3 ways in which the three gamblers have all even. If 3 dices are even, there are 20 ways to do it and only one way in which the three gamblers have all even. 8/42 = 4/21 is the answer. -- by 会员 sdcar2010 (2011/2/17 9:27:30)
有一点点疑问。。
sacar这样算出来貌似是：P(B，D，F三个人是偶数|至少3个人是偶数)
而题目要我们求的是P(B，D，F三个人是偶数)
所以根据条件概率：P(B，D，F三个人是偶数|至少3个人是偶数)还应该乘上P(至少3个人是偶数)，分四种情况则公式是：
P(XY)=P(X1/Y1)*P(Y1)+P(X2/Y2)*P(Y2)+P(X3/Y3)*P(Y3)+P(X4/Y4)*P(Y4) --- (1)
其中：
XY联合概率 -- 题目指定的三个人都拿到偶数的概率X1/Y1 -- 有三个人偶数时，刚好是题目指定那三个人；对应事件Y1 -- 三个人拿到偶数的概率
如果3个dices是even的，那么概率是20/64 ==>(20/64)*(1*1/20)
X2/Y2 -- 有四个人偶数时，刚好包括题目指定的那三个人；对应事件Y2 -- 四个人拿到偶数的概率
如果4个dices是even的，那么概率是15/64 ==>(15/64)*(3*1/15)
X3/Y3 -- 有五个人偶数时，刚好包括题目指定的那三个人；对应事件Y3 -- 五个人拿到偶数的概率
如果5个dices是even的，那么概率是6/64 ==>(6/64)*(3*1/6)
X4/Y4 -- 有六个人偶数时，刚好包括题目指定的那三个人；对应事件Y4 -- 六个人拿到偶数的概率
如果6个dices是even的，那么概率是1/64. ==>(1/64)*(1*1/1)
代入公式(1)
8/64 = 1/8 is the answer.
不知道对不对，欢迎指教。 | 
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发表于 2011-2-17 00:55:52
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发表于 2011-2-18 16:23:23