4 A shipment of 8 television sets contains 2 black-and-white sets and 6color sets. If 2 television sets are to be chosen at random from this shipment, what is the probability that at least 1 of the 2 sets chosen will be a black-and-white set?
Answer:13/28
这个题答案E,我算出来13/28=1-(6*5)/(8*7) 即1-全部是彩电的情况-方法一
(6*2)/(8*7)+1/(8*7)=13/56即一个黑白1个彩电+2各都是黑白的情况-方法二
两个方法的答案应该算出来一样,但是方法二哪里不对了?请帮我指正
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the method 2 should be C(6,1)*C(2,1)/C(8,2)+C(2,2)/C(8,2)
noticing that C(8,2)=8*7/2, you will get the answer after calculating the expression.
谢谢2位,方法二:
一个黑白1个彩电 C(6,1)*C(2,1)/C(8,2)=12/28 这里为什么不是C(6,1)*C(2,1)/C(1,8)*C(1,7)=6/28,我可以当作一个接一个不放回分开取阿?
2各都是黑白的情况C(2,2)/C(8,2)或=C(1,2)*C(1,1)/C(1,8)*C(1,7) =1/28
一个黑白1个彩电 C(6,1)*C(2,1)/C(8,2)=12/28
这里为什么不是C(6,1)*C(2,1)/C(1,8)*C(1,7)=6/28,我可以当作一个接一个不放回分开取阿?
恳请解答!
为什么是[C(2,1)*C(6,1)]/C(8,2)+C(2,2)/C(8,2)
而不是C(6,1)*C(2,1)/C(1,8)*C(1,7)既然分子是一个一个取,分母也应该是一个一个取啊?
现在弄明白了,应该是
1-C(6,1)/C(8,2)=13/28
或者[C(2,1)*C(6,1)]/C(8,2)+C(2,2)/C(8,2)
分母不能是C(1,8)*C(1,7),这样分母里面有重复计算的部分
thanks a lot,
我明白了,分母应该是C28,跟C18C17不一样,后者是分两次分别取
what is the answer?
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