A student moving into a new flat buys a box of 20 light bulbs. Unknown to the student, exactly one of the bulbs is defective. Find:

b) Theconditionalprobability that the second bulb used is not defective,giventhat the first bulb used is not defective;

If the first bulb is not defective then you have 19 bulbs from which you can choose, and one of these is defective so the probability is $\displaystyle \frac{18}{19}$.

This is the easy way to work it out but I am having trouble using the actual formula which may come in handy for other examples:

Let A be the event that the 2nd bulb is not defective.

Let B be the event that the 1st bulb is not defective.

Then $\displaystyle P(A|B)=\frac{P(A\cap B)}{P(B)}$

$\displaystyle P(B) = \frac{19}{20}$ but how can you work out $\displaystyle P(A\cap B))$?

$\displaystyle P(A\cap B) = P(A|B)\cdot P(B)$

$\displaystyle P(A\cap B) = \frac{18}{19}\cdot \frac{19}{20}=\frac{18}{20}$

So how can we work out that $\displaystyle P(A\cap B) = \frac{18}{20}$ from what we are given?

I apologise for any offence my pedantry has caused(Angel)