how do you calculate two events happening in a row
QUOTE=fn686;175735]Suppose n individuals at the Olympics pass a security checkpoint and each is searched with probability P. If exactly x of n individuals are searched, what is the probability two people in a row are searched?[/QUOTE]
I agree with MrF’s first comment and I am not sure that your new statement is any clearer. Let us change to this: “n individuals at the Olympics pass a security checkpoint and exactly x of n individuals are searched, what is the probability two people in a row are searched?”
Clearly $\displaystyle 1 \le x \le \left\lfloor {\frac{n+1}{2}} \right\rfloor $, that is if $\displaystyle n=7$ then $\displaystyle
1 \le x \le 4$ otherwise two would necessarily be searched is a row. We can model this with x 1’s and n-x 0’s. For example, 00101001 means n=8 & x=3 and the third, the fifth and the eighth person was checked. We don’t want 10110000 because that would mean the third and fourth persons were checked. In this case the probability that two will not be checked in a row is $\displaystyle \frac{{_6 C_3 }}{{_8 C_3 }}$. So the probability that two in a row will be checked is $\displaystyle 1 - \frac{{_6 C_3 }}{{_8 C_3 }}$.
So the general answer to your question is $\displaystyle 1 - \frac{{_{n-x+1} C_x }}{{_n C_3 }}$.