Respuesta :
Number of defective light bulbs = 3 and number of good bulbs = 15
So, there are total (3+15) or 18 bulbs in the box.
4 bulbs are chosen from the box without replacement. So, the number of ways for choosing 4 bulbs from total 18 bulbs [tex]= ^1^8C_{4}=3060[/tex]
a. One of the bulbs drawn is good means 1 bulb is chosen from 15 good bulbs and remaining 3 bulbs are chosen from 3 defective bulbs. So, the possible number of ways [tex]= ^1^5C_{1}*^3C_{3} = 15[/tex]
Thus, the probability that one of the bulbs drawn is good [tex]=\frac{15}{3060}=\frac{1}{204}[/tex]
b. None of the bulbs drawn are defective means 4 bulbs are chosen from 15 good bulbs. So, the possible number of ways [tex]=^1^5C_{4}= 1365[/tex]
Thus, the probability that none of the bulbs drawn are defective [tex]= \frac{1365}{3060}=\frac{91}{204}[/tex]
c. At least one of the bulbs drawn are defective means number of defective bulbs can be 1 or 2 or 3.
If 1 bulb is chosen as defective, then the number of ways [tex]=^3C_{1}*^1^5C_{3} = 1365[/tex]
If 2 bulbs are chosen as defective, then the number of ways [tex]=^3C_{2}*^1^5C_{2} = 315[/tex]
and if 3 bulbs are chosen as defective, then the number of ways [tex]=^3C_{3}*^1^5C_{1} = 15[/tex]
Thus, the probability that at least one of the bulbs drawn are defective [tex]=\frac{1365+315+15}{3060}=\frac{1695}{3060}=\frac{113}{204}[/tex]
