# Problem of the Week #40 - December 31st, 2012

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#### Chris L T521

##### Well-known member
Staff member
Thanks to those who participated in last week's POTW!! Here's this week's problem (and the last University POTW of 2012)!

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Problem: For events $E_1$, $E_2,\ldots,\,E_n$, justify the following probability identity: $P(E_1\cap E_2\cap\cdots\cap E_n)=P(E_1)P(E_2\mid E_1)P(E_3\mid E_2\cap E_1)\cdots P(E_n\mid E_1\cap E_2\cap\cdots \cap E_{n-1}).$
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Hint:
Use the fact that $P(A\cap B) = P(A)P(B\mid A)$.

Remember to read the POTW submission guidelines to find out how to submit your answers!

#### Chris L T521

##### Well-known member
Staff member
This week's problem was correctly answered by CaptainBlack and Sudharaka. You can find Sudharaka's solution below:

We shall show this by mathematical induction. When $$n=2$$ the statement is true by the definition of conditional probability. That is,$P(E_1\cap E_2)=P(E_1)P(E_2\mid E_1)$
Suppose that the statement is true for $$n=p\in\mathbb{Z}^{+}$$. That is,
$P(E_1\cap E_2\cap\cdots\cap E_p)=P(E_1)P(E_2\mid E_1)P(E_3\mid E_2\cap E_1)\cdots P(E_p\mid E_1\cap E_2\cap\cdots \cap E_{p-1})~~~~~~~~~~~(1)$
Now consider $$P(E_1\cap E_2\cap\cdots\cap E_p\cap E_{p+1})$$. By the definition of conditional probability we get,
$P(E_1\cap E_2\cap\cdots\cap E_p\cap E_{p+1}) = P(E_1\cap E_2\cap\cdots\cap E_p)P(E_{p+1}\mid E_1\cap E_2\cap\cdots \cap E_{p})~~~~~~~~~~(2)$
By (1) and (2),
$P(E_1\cap E_2\cap\cdots\cap E_{p+1})=P(E_1)P(E_2\mid E_1)P(E_3\mid E_2\cap E_1)\cdots P(E_{p+1}\mid E_1\cap E_2\cap\cdots \cap E_{p})$
Therefore by mathematical induction,
$P(E_1\cap E_2\cap\cdots\cap E_n)=P(E_1)P(E_2\mid E_1)P(E_3\mid E_2\cap E_1)\cdots P(E_n\mid E_1\cap E_2\cap\cdots \cap E_{n-1})\mbox{ for }n\in\mathbb{Z}^+$
Q.E.D.

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