# just wondering

• Nov 12th 2007, 06:11 AM
slevvio
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 $\displaystyle \frac{18}{19}$.

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)
• Nov 12th 2007, 06:33 AM
Plato
$\displaystyle \begin{array}{l} P(X \cap Y) = P(X|Y)P(Y) \\ P(A \cap B) = P(B)P(A|B) = \left( {\frac{{19}}{{20}}} \right)\frac{{18}}{{19}} = \frac{{18}}{{20}} \\ \end{array}$
• Nov 12th 2007, 06:42 AM
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 $\displaystyle P(A\cap B)$?
• Nov 12th 2007, 07:07 AM
Plato
Quote:

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 $\displaystyle P(A\cap B)$?

$\displaystyle P(A \cap B) = \left( {\frac{2}{{20}}} \right)\left( {\frac{1}{{19}}} \right)$
• Nov 12th 2007, 09:55 AM
slevvio
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 $\displaystyle P(A\cap B)$.