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Homework answers / question archive / Probability Consider a seven-game world series between team A and B, where for each game P(A wins)=0

Probability Consider a seven-game world series between team A and B, where for each game P(A wins)=0

Math

Probability

Consider a seven-game world series between team A and B, where for each game
P(A wins)=0.6

a) Find P(A wins series in x game)
b) You hold a ticket for the seventh game. What is the probability that you will get to use it? .answer 0.2765
c) if P(A wins a game)=p, what value of p maximizes your chance in b)?answer p=1/2
d) what is the most likely number of games to be played in the series for p=0.6?
answer x=6

I have the answer but I do not understand the process. Can you explain this with details and step by step What does most likely means?

pur-new-sol

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From the condition, the probability that A wins a game is P(A)=0.6, then the probability that B wins a game is P(B)=1-P(A)=0.4.

(a) If A wins the series in x games, since it is a seven-game series, then A actually wins 4 games. The winner is the one who wins 4 games first. We also note A must win the last game. So A wins 3 games in the first (x-1) games. We select 3 from x-1, and there are C(x-1,3) ways. For each way, it has probability 0.6^3*0.4^(x-1-3)=0.216*0.4^(x-4).
Therefore, P(A wins series in x games)=0.6*C(x-1,3)*0.216*0.4^(x-4)
=0.1296*C(x-1,3)*0.4^(x-4)
(b) If you will get to use the ticket for the seventh game, then in the first 6 games, A wins 3 and B wins 3. There are C(6,3)=6!/(3!)^2=6*5*4/3!=20 different ways for A to win 3 games. For each way, the probability is P(A)^3 * P(B)^3 = 0.6^3 * 0.4^3=0.013824. Thus the probability that you will get to use the ticket for the seventh game is 20*0.013824=0.27648
(c) For a general case, if P(A)=p, then the probability is f(p)=20*p^3*(1-p)^3. Then we have
f(p)=20(p(1-p))^3. To maximize f(p), we only need to maximize p(1-p). We note that p(1-p)<=((p+(1-p))/2)^2=1/4. The equality is obtained only when p=1-p. So p=1/2. When p=1/2, f(p) reaches its maximum value.
(d) At least 4 games are required. At most 7 games are required. Let X be the number of required games, then from (b), we know that P(X=7)=0.27648. Now we compute P(X=4), P(X=5) and P(X=6).
If exactly 4 games are played, then A wins all 4 games or B wins all 4 games. So we have
P(X=4)=0.6^4+0.4^4=0.1552.
If exactly 5 games are played, then A wins 4 games, B wins 1, or A wins 1, B wins 4. All possible cases if A wins finally are : AAABA, AABAA, ABAAA, BAAAA. Similar if B wins finally. So
P(X=5)=4*0.6^4*0.4+4*0.4^4*0.6=0.2688.
If exactly 6 games are played, then
P(X=6)=1-P(X=4)-P(X=5)-p(X=7)=1-0.1552-0.2688-0.27648=0.29952
The most likely number of games is the expected number of the games, which is the expected value of X. So we have
E(X)=4*P(X=4)+5*P(X=5)+6*P(X=6)+7*P(X=7)
=4*0.1552+5*0.2688+6*0.29952+7*0.27648
=5.70 or close to 6.

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