a box contain 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 aand the other will not be defective is 7/15

The given answer is D, but my answer is a .  who can help me to answer it.

1) n/10 * (n-1)/9 =1/15, n*(n-1)=6 ; n=3

2) n/10 * (10-n)/9 =7/15;  n * (10-n)=42; n*2-10n+42=0 ;but B*2-4ac=100-4*1*42<0 , so no answer.

Please discuss it.

should be divided by 2 or 1, not 10 or 9

The probability that one of the bulbs to be drawn will be defective is: n/10 and the probability that the other one of the bulbs to be drawn will not be defective is (10-n)/9, so n/10 * (10-n)/9 =7/15.

n/10 * (10-n)/9 =7/15，你这么计算意味着分两步取灯泡，但题目说是同时取两个灯泡，分两步取还是一步取，结果是不同的。n×(10-n)，代表取一个坏灯泡和一个好灯泡的取法，而同时取两个灯泡的取法等于 10×9×(1/2)＝45,而不是10×9＝90。

所以条件2的算法为 n×（10－n）/45 ＝ 7/15