**Example**

**Example**- 重覆投擲兩粒均勻的骰子, 且設每次投擲的結果彼此不互相影響. 試求
在出現點數和為 7 之前出現點數和為 5 的機率.

*Solution:*- 令
*E*_{n}表示在首*n*-1 次試驗中點數和不為 5 也不為 7, 但第*n*次試驗時點數和為 5 的事件, 則所求的機率為This result may also have been obtained by using conditional probabilities. If we let

*E*be the event that a 5 occurs before a 7, then we can obtain the desired probability,*P*(*E*), by conditioning on the outcome of the first trial, as follows: Let*F*be the event that the first trial results in a 5; let*G*be the event that it results in a 7; and let*H*be the event that the first trial results in neither a 5 nor a 7. Conditioning on which one of these events occurs gives*P*(*E*)=*P*(*E*|*F*)*P*(*F*)+*P*(*E*|*G*)*P*(*G*)+*P*(*E*|*H*)*P*(*H*)orThe reader should note that the answer is quite intuitive. That is, since a 5 occurs on any roll with probability and a 7 with probability , it seems intuitive that the odds that a 5 appears before a 7 should be 6 to 4 against. The probability should be , as indeed it is.

The same argument shows that if*E*and*F*are mutually exclusive events of an experiment, then, when independent trials of this experiment are performed, the event*E*will occur before the event*F*with probability