How to find final probability if I know the probability of the individual events leading to it. I'm trying to understand a homework problem so I'm examining a simpler one. 


*

*The chance of getting heads on a single coin toss is .5.

*The chance of getting heads at least once with 2 coin tosses is
.75, (3/4).

*The chance of getting heads at least twice with 4 coin tosses is
.6875, (11/16).


I got that from simply counting. How would I get the answer without counting, purely with an equation? From the coin toss example, I can see that I can't naively add the .5 chance you have on an individual coin toss for each toss (.5 * n) for n = number of tosses.
What would be the equation to use, with just knowing the probability of the first event and the probability of each successive event, and the number of events? I know this is simple, I can't get it though.
 A: If it is for the events having only two possible outcomes, such as the coin-toss, you can use $$\sum _{r={\frac n2}}^n [(^nC_r)(p)^r(q)^{(n-r)}]$$
where p is the probability of getting the first event, and q is the probability of getting the second event, and $p+q=1$ always. In the coin-toss experiment, we can say that p is the probability of getting heads and q of getting tails. So, $p=\frac12$ and $q=\frac12$.
The above equation is for finding the probability that the event occurs at least $\frac n2$ times, as you gave in your examples. You can change the boundaries of the summation according to the problem.
A: There are two ways to handle "at least" questions. The simpler way is to directly enumerate, as you have done. Alternatively, especially if the "at least" value is small, these problems can be addressed by recognizing that "at least X" is equivalent to "All cases except less than X". For example, what is the probability of getting at least 3 heads on 100 tosses? It is 1 - probability of 0, 1, or two heads.
The probability distribution of $k$ successes in $n$ trials of a yes/no, true/false, heads/tails (Bernoulli random variables) is the binomial distribution.
