Q6:

A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?

A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?

A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?

(1) The probability that the two bulbs to be drawn will be defective is 1/15.

(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.

这道题大家讨论过，讨论帖我也看过了
  由（2）得

Cn,1*C10-n,1/C10,2=7/15=>n=3 or 7 已知，n=3 <5所以n=3

但是没有人用我这种方法算

n/10*10-n/9=7/15 =>n*(10-n)=42 ??? 这种方法得不出n=3，有人能告诉我错在哪里了吗？万分感谢！！！

我刚做到这道题的时候，也排了你这个式子，验算的时候发现，这样算概率的话，有一个隐含的假定是挑选两种bulbs的顺序的一定的，即先挑defective，再挑none-defective，所以漏算了另外一个顺序，即先挑none-defective 再挑defective，后一种选法发生的概率是[(10-n）/10]*[n/9],把两个概率相加，才是7/15

不知道这样解释表达得清楚不清楚

totall agree!