锅中煮有芝麻馅6个,花生馅5个,豆沙馅4个.从中任意舀取4个汤圆,则每种汤圆都至少取到1个的概率为[C(6,2)C(5,1)C(4,1)+C(6,1)C(5,2)C(4,1)+C(6,1)C(5,1)C(4,2)] / C(6+5+4,4)= (300+240+180)/1365= 48/91.C(6,1)C(5,1)C(4,1)C(5,
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锅中煮有芝麻馅6个,花生馅5个,豆沙馅4个.从中任意舀取4个汤圆,则每种汤圆都至少取到1个的概率为[C(6,2)C(5,1)C(4,1)+C(6,1)C(5,2)C(4,1)+C(6,1)C(5,1)C(4,2)] / C(6+5+4,4)= (300+240+180)/1365= 48/91.C(6,1)C(5,1)C(4,1)C(5,
锅中煮有芝麻馅6个,花生馅5个,豆沙馅4个.从中任意舀取4个汤圆,则每种汤圆都至少取到1个的概率为
[C(6,2)C(5,1)C(4,1)+C(6,1)C(5,2)C(4,1)+C(6,1)C(5,1)C(4,2)] / C(6+5+4,4)
= (300+240+180)/1365
= 48/91.
C(6,1)C(5,1)C(4,1)C(5,1)+C(6,1)C(5,1)C(4,1)C(4,1)+C(6,1)C(5,1)C(4,1) C(3,1)] / C(6+5+4,4)
= (300+240+180)/1365
请问我哪里想错了
锅中煮有芝麻馅6个,花生馅5个,豆沙馅4个.从中任意舀取4个汤圆,则每种汤圆都至少取到1个的概率为[C(6,2)C(5,1)C(4,1)+C(6,1)C(5,2)C(4,1)+C(6,1)C(5,1)C(4,2)] / C(6+5+4,4)= (300+240+180)/1365= 48/91.C(6,1)C(5,1)C(4,1)C(5,
你的思路是从三种里面各取一个,然后再在三种里面选取一种取一个,这样取法有重复.举个例子,豆沙馅有4个,我们把它们编号甲乙丙丁,你可能先取到甲,再取的时候取到乙,但也可能先取到乙,再取的时候取到甲,这本来是一种情况,但是你把它们分开取,就成了两种情况.所以,你的答案错了.