1. ## just wondering

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) The conditional probability that the second bulb used is not defective, given that 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 $\frac{18}{19}$.

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

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

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

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

I apologise for any offence my pedantry has caused

2. $\begin{array}{l}
P(X \cap Y) = P(X|Y)P(Y) \\
P(A \cap B) = P(B)P(A|B) = \left( {\frac{{19}}{{20}}} \right)\frac{{18}}{{19}} = \frac{{18}}{{20}} \\
\end{array}$

3. Let me rephrase my question:

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. Let A be the event the 2nd bulb is defective, and let B be the event the 1st bulb is defective. What is $P(A\cap B)$?

4. Originally Posted by slevvio
Let me rephrase my question:

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. Let A be the event the 2nd bulb is defective, and let B be the event the 1st bulb is defective. What is $P(A\cap B)$?
$P(A \cap B) = \left( {\frac{2}{{20}}} \right)\left( {\frac{1}{{19}}} \right)$

5. Thank you but once again I have asked the wrong thing.

A student moving into a new flat buys a box of 20 light bulbs. He takes a number of bulbs from the box without replacement. Unknown to the student, exactly one of the bulbs is defective. Find:

Let A be the event the 2nd bulb is NOT DEFECTIVE and B the event that the 1st bulb is NOT DEFECTIVE. Find $P(A\cap B)$.