The Multiplication Rule 乘法法則

The Multiplication Rule

$P(E_1E_2E_3\ldots E_n)=P(E_1)P(E_2\vert E_1)P(E_3\vert E_1E_2)\ldots P(E_n\vert E_1E_2\ldots E_{n-1})$

Proof:
欲證明乘法法則, 祇需要將條件機率的定義應用到等式的右邊就可以了. 亦即

$\displaystyle P(E_1)\times\frac{P(E_1E_2)}{P(E_1)}\times\frac{P(E_1E_2E_3)}{P(E... ...} \ldots\frac{P(E_1E_2\ldots E_n)}{P(E_1E_2\ldots E_{n-1})}=P(E_1E_2\ldots E_n)$         $\rule[0.02em]{1.0mm}{1.5mm}$

Example
An ordinary deck of 52 playing cards is randomly divided into 4 piles of 13 cards each. Compute the probability that each pile has exactly 1 ace.
Solution:
Define events Ei, i=1,2,3,4 as follows.
E1={ the ace of spades is in any one of the piles }
E2={ the ace of spades and the ace of hearts are in different piles }
E3={ the aces of spades, hearts, and diamonds are all in different piles }
E4={ all 4 aces are in different piles }
The probability desired is P(E1E2E3E4) and by the multiplication rule
P(E_1E_2E_3E_4)=P(E_1)P(E_2|E_1)P(E_3|E_1E_2)P(E_4|E_1E_2E_3)
Now, P(E1)=1 since E1 is the sample space S.
$\displaystyle P(E_2\vert E_1)=\frac{39}{51}$ since the pile containing the ace of spades will receive 12 of the remaining 51 cards.
$\displaystyle P(E_3\vert E_1E_2)=\frac{26}{50}$ since the piles containing the aces of spades and hearts will receive 24 of the remaining 50 cards; and finally, $\displaystyle P(E_4\vert E_1E_2E_3)=\frac{13}{49}$.
Therefore, we obtain that the probability that each pile has exactly 1 ace is
$\displaystyle P(E_1E_2E_3E_4)=\frac{39\cdot 26\cdot 13}{51\cdot 50\cdot 49}\approx .105$
That is, there is approximately a 10.5 percent chance that each pile will contain an ace.          $\rule[0.02em]{1.0mm}{1.5mm}$