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dexar [7]
3 years ago
11

Two teams A and B play a series of games until one team wins three games. We assume that the games are played independently and

that the probability that A wins any game is p. What is the probability that the series lasts exactly four games?
Mathematics
1 answer:
Olenka [21]3 years ago
8 0

Answer:

The probability that the series lasts exactly four games is 3p(1-p)(p^{2} + (1 - p)^{2})

Step-by-step explanation:

For each game, there are only two possible outcomes. Either team A wins, or team A loses. Games are played independently. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

We also need to know a small concept of independent events.

Independent events:

If two events, A and B, are independent, we have that:

P(A \cap B) = P(A)*P(B)

What is the probability that the series lasts exactly four games?

This happens if A wins in 4 games of B wins in 4 games.

Probability of A winning in exactly four games:

In the first two games, A must win 2 of them. Also, A must win the fourth game. So, two independent events:

Event A: A wins two of the first three games.

Event B: A wins the fourth game.

P(A):

A wins any game with probability p. 3 games, so n = 3. We have to find P(A) = P(X = 2).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(A) = P(X = 2) = C_{3,2}.p^{2}.(1-p)^{1} = 3p^{2}(1-p)

P(B):

The probability that A wins any game is p, so P(B) = p.

Probability that A wins in 4:

A and B are independent, so:

P(A4) = P(A)*P(B) = 3p^{2}(1-p)*p = 3p^{3}(1-p)

Probability of B winning in exactly four games:

In the first three games, A must win one and B must win 2. The fourth game must be won by 2. So

Event A: A wins one of the first three.

Event B: B wins the fourth game.

P(A)

P(X = 1).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(A) = P(X = 1) = C_{3,1}.p^{1}.(1-p)^{2} = 3p(1-p)^{2}

P(B)

B wins each game with probability 1 - p, do P(B) = 1 - p.

Probability that B wins in 4:

A and B are independent, so:

P(B4) = P(A)*P(B) = 3p(1-p)^{2}*(1-p) = 3p(1-p)^{3}

Probability that the series lasts exactly four games:

p = P(A4) + P(B4) = 3p^{3}(1-p) + 3p(1-p)^{3} = 3p(1-p)(p^{2} + (1 - p)^{2})

The probability that the series lasts exactly four games is 3p(1-p)(p^{2} + (1 - p)^{2})

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