Question

Question 36
A carton of 24 bulbs contains 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?

Answer 1

$\because$ Total number of bulbs , n(S)= 24

Let $E_1$ = Event of selecting not defective bulb = Event of selecting good bulbs
$n\left(E_1\right)=18$
$\therefore P\left(E_1\right)=\frac{n\left(E_1\right)}{n\left(S\right)}=\frac{18}{24}=\frac{3}{4}$
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)$=\frac{5}{23}$