You have many other postings. So you should understand that this is not a homework service. Please either post some of your own work on this problem or explain what you do not understand about the question.
A computer game stimulate a match beteewn A and B. The match consist of at most 3 sets. Each set is won by either A or B and the match is won by 1st player to win the two sets.
The stimulation uses the following rules:
-The probability that A wins the 1st set is 0.6
-For each set after the 1st, the conditional probability that A wins that set, given that A won the preceding set, is 0.7
-For each set after the 1st, the cnditional probability that B wins that set, given B won the preceding set, is 0.8
Find the probability that (i)A wins the second set
(ii)A wins the match
Conceptually this problem is straightforward if you set it up right in your head (or on paper, a binary tree of probabilities helps). Define a binary game (A, B) to be like one of the sets above. Only one of A or B can win with some specified probability. Thus, set one can be denoted by (0.6, 0.4). Clearly, the probability of the game sums to 1 and we know the right-hand term since we're given the left-hand term by Pr(A in first set) = 0.6. The real crux of this problem, that if you understand it makes the rest of the problem easy, is that it consists in a number of these games. For instance, we can only have another game (A, B) in set 2, but there are two different ways we can get to this game. The first path is if A won in the first set. Thus, let us denote this game A(A, B) and the alternative B(A, B). We're given Pr(A in second set | A in first set) = 0.7. So, A(0.7, B) is that game and similarly B(A, 0.8) is the second game (fill in the rest). If we want to know what a given probability is, like for question (i) we have to look for Pr(A in second) in terms of how many ways it can happen. Clearly, it can happen in two of the games. Namely, Pr("A in second & A in first" OR "A in second & B in first"). But notice that Pr(A2 & A1) = Pr(A2 | A1) * Pr(A1) by definition of conditional probability (here '&' can also denote set intersection). Thus, for each of the games A(A, B) and B(A, B), the games sum to 1, but for the entire second set we require that the two games jointly sum to unity. Does this happen? The answer is yes, based on the above equation. For instance Pr(A2 & A1) = Pr(A2 | A1) * Pr(A1) = 0.7 * 0.6 = 4.2. With Pr(A2 & B1) = Pr(A2 | B1) * Pr(B1) = 0.2 * 0.4 = 0.8 we can see that the answer to (i) is 1/2. Why is Pr(A2 | B1) = 0.2? Because this is from the game B(A, B) = B(A, 0.8). Thus, A in this game has to be 1 - 0.8 = 0.2, just like in the game for set 1. Set 3 will consists of four such games with 8 outcomes total. The complexity rises, but the calculations are just as simple. In fact, the sub-games will consists of the same probabilities as for set 2 given the rules provided. So calculating any probability is just a conjunction of each node in the path of the game as derived by the conditional probability equation used above (e.g., Pr(A3 | A2 & B1) = 0.4 * 0.7 * 0.7 = 0.196.
The answer to (ii) is greater than that for (i) but less than 0.75 by my calculations.
Actually, I messed up on my calculation last night. I do get the same answer, though. Let me walk you through it in detail. Here we need to know in how many possible ways can have A win the match. Using concatenation to describe these events, there are three events we need to know: AA, BAA, ABA. Let us analyze them in turn.
AA is the case in which A wins the second set and A wins the first set. Thus, we are after . As explained earlier, the conditional probability is just the probability of the game at that set. Thus, on the notation used, . What we are after is the value of A inside of the game A(A, B). We are given that this game is A(0.7, 0.3), the right-term being simply the first given taken from 1. Therefore,
BAA proceeds in a similar fashion, but now we have to consider the third level games. In particular, we have to look at . For notational purposes the "A2.B1" means "A2 and B1." Why is this the probabilities we require? Because . Let us see what we're given for each set. . This one is clearly established. However, , and we're given the right-term. Thus, we can define that game as . Lastly, we need to know the third game. Given the definitions, it is entirely defined in terms of its previous set. The games are, in fact, identical in the second set. If we're conditional on a prior A set, then we look at ; otherwise, we look at . Therefore,
In similar analysis, . Summing the three individual probabilities does yield 51.2%
I have altered the notation a little from my first go around. I think an easier way to think of the conditional probabilities is to make the "given" parameter either 'A' or 'B', since that is all the information that is required. The only probabilities we need to know are the first set . The conditional games are given by their "given" parameter: or . We can then replace the 'X' in those statements by either 'L' or 'R' to indicate whether it is the left or right term in the given game. This gives us a standard way to express the A or B instead of the 'A1' or 'B2' utilized before.
In this notation, the third set can just as easily be put into terms of these conditional games. We can draw it as a three-tier "tree" using these games to define the nodes. To calculate the probability at any node you simply multiply each node along the path, as was the case in the calculations above. In this notation, for instance,
Here clearly 'L' refers to 'A' in a given game and 'R' to 'B' in a given game. I think this simplified notation should help simplify the problem and its solutions.