Probability of winning a best of 3 out of 5 game Two people play a series of independent games. Person 1's probability of winning any game is 0.6, which leaves 0.4 for Person 2. If they play a best of 5 tournament (3/5), find the following. 
(1)Probability that person 1 wins the tournament in 3 games. 
(2) Probability that the tournament lasts exactly 3 games. 
(3) probability that the tournament lasts exactly 4 games. 
(4) Probability that Person 1 wins the tournament. 
(5) Probability that it lasts only 3 games if won by person 1. 
(6) Probability that person 1 won the tournament if it lasted exactly 3 games. 
So, I'm pretty lost on how to even start with these problems. 
I was thinking that for (1) it would be (3/5)*(3/5) -- Person1's probability of winning times the best of 5 wins. Is that even on the right track? 
And how would you start the others?
 A: $1.$ This is easy, the fact you didn't answer it correctly is probably due to not understanding the question. Person $1$, let's call her Alice, wins the tournament in $3$ games if she wins Games $1$, $2$, and $3$. We are assuming independence, so the probability is $(0.6)^3$.
$2.$ The tournament lasts exactly $3$ games if Alice wins Games $1$, $2$, and $3$, or Betty does. The probability Betty does is $(0.4)^3$, so our required probability is $(0.6)^3+(0.4)^3$.   
$3.$ This is more complicated. The tournament lasts exactly $4$ games if (i) Alice wins the $4$th game, and exactly $2$ of the other $3$ or (ii) Betty wins the $4$th game, and exactly $2$ of the other $3$. 
For (i), winning $2$ of the first $3$ can happen in the patterns WWL, WLW, and LWW. Each of these has probability $(0.6)^2(0.4)$. Multiply by $3$ because of the $3$ different ways. We get $3(0.6)^2(0.4)$. Multiply by the probability Alice wins the $4$th game. We get $3(0.6)^3(0.4)$.
We get a similar expression for (ii), reversing the roles of $0.6$ and $0.4$. Add: we get
$$3(0.6)^3(0.4)+3(0.4)^3(0.6).$$
$4.$ This is a sum of the probabilities that Alice wins in $3$, in $4$, and in $5$. We already know the answers to the first two: $(0.6)^3$ and $3(0.6)^3(0.4)$ respectively. I will leave to you to find the probability Alice wins in $5$. Hint: She has to win the $5$th game, and exactly $2$ of the first $4$.
$5.$ Hint: It has something to do with the answers to $1$ and $4$. The key word is conditional probability. 
In symbols, let $A$ be the event "Tournament lasts $3$ games" and
$B$ the event "Alice wins tournament." We want $\Pr(A|B)$.
$6.$ Again, a conditional probability. 
After you have worked on the problems for a while, perhaps I can add to the hints. Would need to know what you have been exposed to about conditional probability.
Remark: In case you are not familiar with the tournament setup, here is an explanation of how it works. As soon as one person has won $3$ games, the tournament is over. So the tournament can last $3$, $4$, or $5$ games. If some person wins Games $1$, $2$, and $3$, the tournament is over, no more games are played.  
