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！！！ | 
