PLEASE help me guys!!

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Originally Posted by Kiwigirl

I'm in desperate need of guidance. I have no idea how to answer this question and it is so frustrating for me because I've spent so long trying to figure it out! Thank you for your help :)

In a tournament, two players continue to play each other until one of them wins three games. If the probability that A wins any single game is 0.4 and the probability that B wins it is 0.5, find

a) the probability the first game is a draw

The probability that the first game is a draw is the probability that
neither wins, and since A wins with probability 0,4 and B wins with
probability 0.5, the probability of a draw on any game is 1-0.4-0.5=0.1.

Therefore the probability that the first game is a draw is the probability
that any game is a draw and is equal to 0.1

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b) the probability A wins the first three games

The results of each of the first three games are independent, and the
probability of A winning in any one of them is 0.4 so the probability of
A winning the first three games is the product of the probabilities of
A winning each of them=0.4x0.4x0.4=0.064

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c) the probability the tournament is just three games long

There are two ways that the tournament can be just three games long;
A can win the first three games, or B can win the first three games. The
probabilities of these outcomes are 0.4x0.4x0.4 and 0.5x0.5x0.5
respectively, and these outcomes are independent the probability of
one or the other occurring is the sum of their individual probabilities and
so is:

0.4x0.4x0.4 + 0.5x0.5x0.5 = 0.189

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d) the probability A wins the first two games, and not the third.

As the results of each game are independent this is the product of the
probabilities of the required outcomes of each game. The probability that
A wins a game is 0.4, the probability that A does not win a game is 1-0.4=0.6.
So the required probability is 0.4x0.4x0.6=0.096

RonL