A carton of 24 bulbs contain 6 defective bulbs. One bulb is drawn at random. What is the probability that the bulb is not defective? If the bulb selected is
defective and it is not replaced and a second bulb is selected at random from the rest, what is the probability that the second bulb is defective?
∴Total number of bulbs, n (S) = 24
Let $E_{1}=$ Event of selecting not defective bulb $=$ Event of selecting good bulbs
$n\left(E_{1}\right)=18$
$\therefore$ $P\left(E_{1}\right)=\frac{n\left(E_{1}\right)}{n(S)}=\frac{18}{24}=\frac{3}{4}$
Suppose, the selected bulb is defective and not replaced, then total number of bulbs remains in a carton, $n(S)=23$.
In them, 18 are good bulbs and 5 are defective bulbs.
$\therefore P$ (selecting second defective bulb) $=\frac{5}{23}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.