Need help finding the probability I have problem where you are given the number of times you must win, and the chance of winning a game. I need to find the probability that I win all 4 times.
Example: I am only given two inputs, 4 (the number of times I must win) and 0.4 (the chance of winning each time).
I would expect the answer to just be 4 * 0.4, but the answer that is correct for this scenario is 0.290 (rounded off). I've tried multiple methods to achieve this answer, but I cannot for the the life of me get it.
 A: You said you want "P(Win on all four trials)". 
You use the fact that the trials are independent to compute the joint probability.
That is, you use the rule $P(A\cap B)=P(A)P(B)$ (for independent events A and B). You'll need to apply that rule recursively to get the equivalent for four events.
See here

Interpreting the question to be 
"In a series consisting of a maximum f 7 games, the only possible outcomes on a game are win (W) or lose (L) -- the series is won as soon as one side wins a majority of - games (i.e. 4 wins, making them the winner). Our side has a 0.4 chance to win each game, and the outcomes of each game are independent of all the other games. What is the chance our team wins the series?"
Then you need to consider the mutually exclusive cases 
(i) The series is won in exactly 4 games (4W, 0L)
 (ii) The series is won in exactly 5 games (4W, 1L)
(iii) The series is won in exactly 6 games (4W, 2L)
 (iv) The series is won in exactly 7 games (4W, 3L)  
and solve each separately. Within each you have a simpl binomial problem.
-- but take care! note that in each case the last game must be a win for our side, so that's not part of the possible arrangements of the W/L ordering within each case.
A: Looking at the question you posted, it looks like you need the probability of winning 4 games before the opposing team wins 4 games. Then there are four possibilities: your team wins the first four, your team wins 3 of the first 4 and the last game, your team wins 3 of the first 5 and the last game, your team wins 3 of the first 6 and the last game. If you add up those probabilities, you should get about .2898. 
