We will be using the fact that [tex] P(A \, and \,B) = P(A) \cdot P(B) [/tex], where [tex] P(A) [/tex] is the probability that the first bulb works and [tex] P(B) [/tex] is the probability that the second bulb works.
The probability that the first bulb works is [tex] \frac{20}{24} = \frac{5}{6} [/tex]. However, when we take one out (given that the first bulb works) we now have 19 working bulbs and 4 bad ones. This means that the probability that the second bulb works is [tex] \frac{19}{23} [/tex].
Now, we can find our final probability:
[tex] \dfrac{5}{6} \cdot \dfrac{19}{23} = \boxed{\dfrac{95}{138}} [/tex]
